5 


A.      S.      Hallid  IE, 


SAN  FRANCISCO. 


tiPfffiNN 


■■I 


/ 


ROBINSON'S  MATHEMATICAL  SERIES 


ELEMENTS 


GEOMETRY, 


PLANE  AND  SPHERICAL  TRIGONOMETRY; 


WITH 


NUMEROUS  PRACTICAL  PROBLEMS 

BY 

HOEATIO  K  BOBINSON,  LL.  D., 

AUTHOR  OP  A  FUIX  COURSE  OF  MATHEMATICS. 

OF  THE 


OF 


NEW    YORK: 

IVISON,  PHINNEY  &  CO.,  48  AND  50  WALKER  ST. 

CHICAGO :  S.  C.  GRIGGS  &  CO.,  39  AND  41  LAKE  ST. 

BOSTON  :    BROWN    A   TAGGARD.      PHILADELPHIA  :    SOWER,  BARNES    k   CO.,  AND 

J.  B.  LIPPINCOTT    &    CO.      CINCINNATI  :    MOORE,  WILSTACH,  KEYS    &    CO. 

SAVANNAH  :    J.  M.  COOPER    &   CoT      ST.  LOUIS  :    KEITH    4;    WOODS. 

NEW  ORLEANS  :  E.  R.  STEVENS  &  CO.      DETROIT:   RAYMOND 

&   LAPHAM.      BALTIMORE  :   CUSHING   &   BAILEY. 


1860 


Robinson's  Complete  Mathematical  Course. 


•*--*- 


ROBINSON'S  SYSTEM   OP   MATHEMATICS, 

Recently  revised  and  enlarged,  is  now  the  most  extensive,  complete,  practical 

and  scientific  Mathematical  Series  published  in  this  country. 

»•  i    —    i   » 

1.  Robinson's  Progressive  Primary  Arithmetic.    Illustrated.    $0  15 

2.  Robinson's  Progressive  Intellectual  Arithmetic,  for  ad- 
•    vanced  Classes,  with  an  Original  and  Comprehensive  System 

of  Analysis 0  25 

3.  Robinson's  Progressive  Practical  Arithmetic;  a  complete 

work  for  Common  Schools  and  Academies.        .        .        .  0  56 

4.  Key  to  Robinson's  Progressive  Practical  Arithmetic.         .      0  50 

5.  Robinson's  Progressive  Higher  Arithmetic.         .       .  0  75 

6.  Key  to  Robinson's  Progressive  Higher  Arithmetic.    .  0  75 

7.  Robinson's  New  Elementary  Algebra:  a  clear  and  simple 
Treatise  for  Beginners 0  75 

8.  Key  to  Robinson's  New  Elementary  Algebra.      .        .  0  75 

9.  Robinson's  University  Algebra :  a  full  and  complete  Trea- 
tise for  Academies  and  Colleges 1  25 

10.  Key  to  Robinson's  University  Algebra ;  separate.        .       .100 

11.  Robinson's  Geometry  and  Trigonometry ;  with  applications 

to  practical  examples .       1  50 

12.  Robinson's  Surveying  and  Navigation;  combining  theory 

with  practice 1  50 

13.  Robinson's  Analytical  Geometry  and  Conic  Sections ;  made 

clear  and  comprehensive  to  common  minds 1  50 

14.  Robinson's  Differential  and  Integral  Calculus ;  a  full  and 
complete  Treatise 1  50 

15.  Robinson's  Elementary  Astronomy;  designed  to  teach  the 

first  principles  of  this  Science 0  75 

16.  Robinson's  University  Astronomy;  for  advanced  classes  in 
Academies  and  Colleges 1  75 

17.  Robinson's  Concise  Mathematical  Operations :  a  book  of  re- 
ference for  the  Teacher,  embracing  the  gems  of  Mathematical 
Science 2  25 

18.  Key  to  Robinson's  University  Algebra,  Geometry,  Survey- 
ing, and  Calculus ;  in  1  vol 1  50 

Entered,  according  to  Act  of  Congress,  in  the  year  1860,  by 

H.  N.  ROBINSON,  LL.D., 

in  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Northern  District 

of  New  York. 


JOHN  PAGAN,  8TERE0TTPEE,   PHILADELPHIA. 


PREFACE. 


In  the  preparation  of  this  work,  the  Author's  previous  treatise, 
"Elements  of  Geometry,  Plane  and  Spherical  Trigonometry,  and  Conic 
Sections,"  has  formed  the  ground-work  of  construction.  But  in 
adapting  the  work  to  the  present  advanced  state  of  Mathematical  edu- 
cation in  our  best  Institutions,  it  was  found  necessary  to  so  alter  the 
plan,  and  the  arrangement  of  subjects,  as  to  make  this  essentially  a 
new  work.  The  demonstrations  of  propositions  have  undergone  radical 
changes,  many  new  propositions  have  been  introduced,  and  the  number 
of  Practical  Problems  greatly  increased,  so  that  the  work  is  now  be- 
lieved to  be  as  full  and  complete  as  could  be  desired  in  an  elementary 
treatise. 

In  view  of  the  fact  that  the  Seventh  Book  is  so  much  larger  than  the 
others,  it  may  be  asked  why  it  is  not  divided  into  two  ?  We  answer, 
that  classifications  and  divisions  are  based  upon  differences,  and  that 
the  differences  seized  upon  for  this  purpose  must  be  determined  by  the 
nature  of  the  properties  and  relations  we  wish  to  investigate.  There 
is  such  a  close  resemblance  between  the  geometrical  properties  of  the 
polyedrons  and  the  round  bodies,  and  the  demonstrations  relating  to 
the  former  require  such  slight  modifications  to  become  applicable  to  the 
latter,  that  there  seems  no  sufficient  reason  for  separating  into  two 
Books  that  part  of  Geometry  which  treats  of  them. 

The  subject  of  Spherical  Geometry,  which  has  been  much  extended 
in  the  present  edition,  is  placed  as  before,  as  an  introduction  to  Spheri- 
cal Trigonometry.  The  propriety  of  this  arrangement  may  be  ques- 
tioned by  some  ;  but  it  is  believed  that  much  of  the  difficulty  which  the 
student  meets  in  mastering  the  propositions  of  Spherical  Trigonometry, 
arises  from  the  fact  that  he  is  not  sufficiently  familiar  witl^the  geome- 
try of  the  surface  of  the  sphere  ;  and  that,  by  having  the  propositions 
of  Spherical  Geometry  fresh  in  his  mind  when  he  begins  the  study  of 
Spherical  Trigonometry,  he  will  be  as  little  embarrassed  with  it  as 
with  Plane  Trigonometry. 

(Hi) 

1 n^a^i 


iv  PREFACE. 

Both  author  and  teacher  must  yield  to  the  demands  of  the  age,  and 
by  a  judicious  combination  of  the  abstract  and  the  concrete,  the  theo- 
retical and  the  practical,  make  the  student  feel  that  what  he  learns 
with  perhaps  painful  effort  at  first,  may  be  made  available  in  import- 
ant applications. 

In  teaching  Geometry  and  Trigonometry,  questions  should  be  asked, 
extra  problems  given,  and  original  demonstrations  required  when  the 
proper  occasions  arise ;  but  care  should  be  taken  that  the  pupil's  powers 
are  not  over-tasked.  By  helping  him  through  his  difficulties  in  such 
a  way  that  he  shall  be  scarcely  conscious  of  having  received  assistance, 
he  will  be  encouraged  to  make  new  and  greater  efforts,  and  will  finally 
acquire  a  fondness  for  a  study  that  may  have  been  highly  repugnant 
to  him  in  the  beginning. 

A  demonstration  that  is  easily  followed  and  comprehended  by  one, 
may  be  obscure  and  difficult  to  another ;  hence  the  advantage  that  will 
sometimes  be  gained  by  giving  two  or  more  demonstrations  of  the  same 
proposition.  When  the  student  perceives  that  the  same  results  may 
frequently  be  reached  by  processes  entirely  different,  he  will  be  stimu- 
lated to  independent  exertion,  and  in  no  respect  can  the  teacher  better 
exhibit  his  tact  than  in  directing  and  encouraging  such  efforts. 

Instances  will  be  found  throughout  the  work  in  which  the  more  im- 
portant propositions  are  twice  and  three  times  demonstrated ;  and  as 
the  methods  of  demonstration  are  in  each  case  quite  different,  it  is 
believed  that  extra  space  has  not  been  thus  occupied  unprofitably. 

Practical  rules  with  applications  will  be  found  throughout  the  work, 
and  in  addition  to  these,  there  are  in  both  the  Geometry  and  the  Trigo- 
nometry, full  collections  of  carefully  selected  Practical  Problems. 
These  are  given  to  exercise  the  powers  and  test  the  proficiency  of  the 
pupil,  and  when  he  has  mastered  the  most  or  all  of  them,  it  is  not 
likely  that  he  will  rest  satisfied  with  present  acquisition,  but  conscious 
of  augmented  strength  and  certain  of  reward,  he  will  enter  new  fields  of 
investigation. 

The  Author  has  been  aided,  in  the  preparation  of  the  present  work, 
by  J.  F.  Quinby,  A.  M.,  of  the  University  of  Rochester,  N.  Y.,  late 
Professor  of  Mathematics  in  the  United  States  Military  Academy  at 
West  Point,  and  J.  H.  French,  LL.  D.,  of  Syracuse,  New  York.  The 
thorough  Scholarship,  and  long  and  successful  experience  of  these  gen- 
tlemen in  $ie  class-room,  rendered  them  eminently  qualified  for  the 
task ;  and  to  them  the  public  are  indebted  for  much  that  is  valuable, 
both  in  the  matter  and  arrangement  of  this  treatise. 

October,  1860. 


CONTENTS. 


PLANE    GEOMETRY. 

DEFINITIONS. 

Geometrical  Magnitudes Page  9 

Plane  Angles 10 

Plane  Figures  of  Three  Sides 12 

Plane  Figures  of  Four  Sides 13 

The  Circle 14 

Units  of  Measure 15 

Explanation  of  Terms .' 16 

Postulates 16 

Axioms 17 

Abbreviations , 17 

BOOK  I. 
Of  Straight  Lines,  Angles,  and  Polygons 19 

BOOK  II. 

Proportion,  and  its  Application  to  Geometrical  Investigations. ...     59 

BOOK  III. 

Of  the  Circle,  and  the  Investigation  of  Theorems  dependent  on  its 

Properties 88 

1*  (v) 


vi  CONTENTS. 

BOOK  IV. 

Problems  in  the  Construction  of  Figures  in  Plane  Geometry Ill 

BOOK  Y. 
On  the  Proportionalities  and  Measurement  of  Polygons  and  Circles.  130 
Practical  Problems 142 

BOOK  VI. 

On  the  Intersections  of  Planes,  the  Relative  Positions  of  Planes, 
■     and  of  Planes  and  Lines 152 

BOOK  VII, 

Solid  Geometry 172 

Practical  Problems 229 

BOOK  VIII. 

Practical  Geometry.  —  Application  of  Algebra  to  Geometry,  and 

also  Propositions  for  Original  Investigation 231 

Miscellaneous  Propositions  in  Plane  Geometry 238 


TRIGONOMETRY. 

PART  I. 

\ 

PLANE   TRIGONOMETRY. 

SECTION  I. 

Elementary  Principles 244 

Definitions 245 

Propositions 248 

Equations  for  the  Sines  of  the  Angles 260 

Natural  Sines,  Cosines,  etc 265 

Trigonometrical  Lines  for  Arcs  exceeding  90° 270 


CONTENTS.  vii 

SECTION  II. 

Plane  Trigonometry,  Practically  Applied 272 

Logarithms ; 278 

GENERAL  APPLICATIONS  WITH  THE  USE  OF  LOGARITHMS. 

I.  Eight- Angled  Trigonometry 288 

II.  Oblique- Angled  Trigonometry 291 

Practical  Problems 295 


SECTION  III. 

Application  of  Trigonometry  to  Measuring  Heights  and  Distances.  298 
Practical  Problems 305 


PART  II. 

SPHERICAL  GEOMETRY  AND  TRIGONOMETRY. 

SECTION  I. 
Spherical  Geometry 310 


SECTION  II. 

Right- Angled  Spherical  Trigonometry 330 

Napier's  Circular  Parts 335 


•  SECTION  III. 

Oblique- Angled  Spherical  Trigonometry 337 

Napier's  Analogies 343 


viii  CONTENTS. 

SECTION  IV. 

Spherical   Trigonometry  Applied.  —  Solution  of  Right- Angled 

Spherical  Triangles 353 

Practical  Problems 35G 

Solution  of  Quadrantal  Triangles 358 

Practical  Problems 361 

Solution  of  Oblique-Angled  Spherical  Triangles 362 

Practical  Problems 367 

SECTION  V. 

Spherical  Trigonometry  applied  to  Astronomy 370 

Application  of  Oblique- Angled  Spherical  Triangles 373 

Spherical  Trigonometry  applied  to  Geography 377 

Table  of  Mean  Time  at  Greenwich 379 

SECTION  VI. 
Regular  Polyedrons 380 


^^      or  the    ^ 


GEOMETEY. 


DEFINITIONS. 

1.  Geometry  is  the  science  which  treats  of  position,  and 
of  the  forms,  measurements,  mutual  relations,  and  pro- 
perties of  limited  portions  of  space. 

Space  extends  without  limit  in  all  directions,  and  contains  all 
bodies. 

2.  A  Point  is  mere  position,  and  has  no  magnitude. 

3.  Extension  is  a  term  employed  to  denote  that  pro- 
perty of  bodies  by  virtue  of  which  they  occupy  definite 
portions  of  space.  The  dimensions  of  extension  are 
length,  breadth,  and  thickness. 

4.  A  Line  is  that  which  has  extension  in  length  only. 
The  extremities  of  a  line  are  points. 

5.  A  Eight  or  Straight  Line  is  one  all  of  whose  parts 
lie  in  the  same  direction. 

6.  A  Curved  Line  is  one  whose  consecutive  parts,  how- 
ever small,  do  not  lie  in  the  same  direction. 

7.  A  Broken  or  Crooked  Line  is 
composed  of  several  straight  lines, 
joined  one  to  another  successively, 
and  extending  in  different  directions. 

When  the  word  line  is  used,  a  straight  line  is  to  be  understood, 
unless  otherwise^  expressed. 

8.  A  Surface  or  Superficies  is  that  which  has  extension 
in  length  and  breadth  only. 

9.  A  Plane  Surface,  or  a  Plane,  is  a  surface  such  that 

m 


10  GEOMETRY. 

if  any  two  of  its  points  be  joined  by  a  straight  line,  every 
point  of  this  line  will  lie  in  the  surface. 

10.  A  Curved  Surface  is  one  which  is  neither  a  plane, 
nor  composed  of  plane  surfaces. 

U.  A  Plane  Angle,  or  simply  an  Angle, 
is  the  difference  in  the  direction  of  two 
lines  proceeding  from  the  same  point. 

The  other  angles  treated  of  in  geometry  will  be  named  and  defined 
in  their  proper  connections. 

12.  A  Volume,  Solid,  or  Body,  is  that  which  has  exten- 
sion in  length,  breadth,  and  thickness. 

These  terms  are  used  in  a  sense  purely  abstract,  to  denote  mere 
space  —  whether  occupied  by  matter  or  not,  being  a  question  with 
which  geometry  is  not  concerned. 

Lines,  Surfaces,  Angles,  and  Volumes  constitute  the 
different  kinds  of  quantity  called  geometrical  magnitudes. 

13.  Parallel  Lines  are  lines  which  have  

the  same  direction. 

Hence  parallel  lines  can  never  meet,  however  far  they  may  be 
produced;  for  two  lines  taking  the  same  direction  cannot  approach 
or  recede  from  each  other. 

Two  parallel  lines  cannot  be  drawn  from  the  same  point;  for  if 
parallel,  they  must  coincide  and  form  one  line. 

PLANE    ANGLES. 

To  make  an  angle  apparent,  the  two 
lines  must  meet  in  a  point,  as  AB  and 
-4(7,  which  meet  in  the  point  A. 

Angles  are  measured  by  degrees. 

14.  A  Degree  is  one  of  the  three  hundred  and  sixty 
equal  parts  of  the  space  about  a  point  in  a  plane. 

If,  in  the  above  figure,  we  suppose  A  C  to  coincide  with  AB, 
there  will  be  but  one  line,  and  no  angle;  but  if  AB  retain  its  posi- 
tion, arid  A  C  begin  to  revolve  about  the  point  A,  an  angle  will  be 
formed,  an<J  its  magnitude  will  be  expressed  by  that  number  of  the 


DEFINITIONS.  11 

360  equal  spaces  about  the  point  A,  which  is  contained  between 
AB  and  AC. 

Angles  are  distinguished  in  respect  to  magnitude  by 
the  terms  Eight,  Acute,  and  Obtuse  Angles. 

15.  A  Right  Angle  is  that  formed  by  one 
line  meeting  another,  so  as  to  make  equal 
angles  with  that  other. 

The  lines  forming  a  right  angle  are  perpendicular 
to  each  other. 

16.  An  Acute  Angle  is  less  than  a  right 
angle. 

17.  An  Obtuse  Angle  is  greater  than 
a  right  angle.  N. 

Obtuse  and  acute  angles  are  also  called 
oblique  angles;  and  lines  which  are  neither  parallel  nor  perpen- 
dicular to  each  other  are  called  oblique  lines. 

18.  The  Vertex  or  Apex  of  an  angle  is  the  point  in  which 
the  including  lines  meet. 

19.  An  angle  is  commonly  designated  by  a  letter  at  its 
vertex;  but  when  two  or  more  angles  have  their  vertices 
at  the  same  point,  they  cannot  be 
thus  distinguished. 

For  example,  when  the  three  lines 
AB,  A  C,  and  AD  meet  in  the  common 
point  A,  we  designate  either  of  the  an- 
gles formed,  by  three  letters,  placing 
that  at  the  vertex  between  those  at  the 
opposite  extremities  of  the  including 
lines.  Thus,  we  say,  the  angle  BAG, 
etc.  B 

20.  Complements.  —  Two  angles  are  said  to  be  comple- 
ments of  each  other,  when  their  sum  is  equal  to  one  right 
angle. 

21.  Supplements.  —  Two  angles  are  said  to  be  supple- 
ments of  each  other,  when  their  sum  is  equal  to  two  right 
angles. 


12 


GEOMETRY. 


PLANE    FIGURES. 

22.  A  Plane  Figure,  in  geometry,  is  a  portion  of  a 
plane  bounded  by  straight  or  curved  lines,  or  by  both 
combined. 

23.  A  Polygon  is  a  plane  figure  bounded  by  straight 
lines,  called  the  sides  of  the  polygon. 

The  least  number  of  sides  that  can  bound  a  polygon  is 
three,  and  by  the  figure  thus  bounded  all  other  polygons 
are  analyzed. 

FIGURES     OF    THREE    SIDES. 

24.  A  Triangle  is  a  polygon  having  three  sides  and 
three  angles. 

Tri  is  a  Latin  prefix  signifying  three;  hence  a  Triangle  is  lite- 
rally a  figure  containing  three  angles.  Triangles  are  denominated 
from  the  relations  both  of  their  sides  and  angles. 

25.  A  Scalene  Triangle  is  one  in 
which  no  two  sides  are  equal. 


26.  An  Isosceles  Triangle   is  one  in 
which  two  of  the  sides  are  equal. 


27.  An  Equilateral  Triangle  is  one  in 
which  the  three  sides  are  equal. 


28.  A  Right -Angled  Triangle  is  one 
which  has  one  of  the  angles  a  right 


angle, 


29.  An  Obtuse- Angled  Triangle  is  one 
having  an  obtuse  angle. 


DEFINITIONS.  13 


30.  An  Acute-Angled  Triangle  is  one 
in  which  each  angle  is  acute. 


31.  An  Equiangular  Triangle   is   one 

having  its  three  angles  equal. 


Equiangular  triangles  are  also  equilateral,  and  vice  versa. 
FIGURES    OF    FOUR    SIDES.     • 

32.  A  Quadrilateral  is  a  polygon  having  four  sides  and 
four  angles. 

33.  A  Parallelogram  is  a  quadrilateral  -, 

which  has  its  opposite  sides  parallel.  /  / 

Parallelograms  are  denominated  from  the  rela- ' 

tions  both  of  their  sides  and  angles. 

34.  A  Rectangle  is  a  parallelogram  hav- 
ing its  angles  right  angles. 


35.  A  Square  is  an  equilateral  rectangle. 

36.  A   Rhomboid  is  an  oblique-angled  parallelogram. 


37.  A  Rhombus  is  an  equilateral  rhom- 
boid. 


38.  A  trapezium  is  a  quadrilateral  having 
no  two  sides  parallel. 

39.  A  Trapezoid  is  a  quadrilateral  in 
which  two  opposite  sides  are  parallel,  and 
the  other  two  oblique. 

40.  Polygons  bounded  by  a  greater  number  of  sides 
2 


14  GEOMETRY, 

than  four  are  denominated  only  by  the  number  of  sides. 
A  polygon  of  ^ve  sides  is  called  a  Pentagon,  of  six  a 
Hexagon,  of  seven  a  Heptagon,  of  eight  an  Octagon,  of 
nine  a  Nonagon,  etc. 

41.  Diagonals  of  a  polygon  are  lines 
joining  the  vertices  of  angles  not  ad- 
jacent. 

42.  The  Perimeter  of  a  polygon  is  its  boundary  consid 
ered  as  a  whole. 

43.  The  Base  of  a  polygon  is  the  side  upon  which  the 
polygon  is  supposed  to  stand. 

44.  The  Altitude  of  a  polygon,  is  the  perpendicular 
distance  between  the  base  and  a  side  or  angle  opposite 
the  base. 

45.  Equal  Magnitudes  are  those  which  are  not  only 
equal  in  all  their  parts,  but  which  also,  when  applied  the 
one  to  the  other,  will  coincide  throughout  their  whole 
extent. 

46.  Equivalent  Magnitudes  are  those  which,  though  they 
do  not  admit  of  coincidence  when  applied  the  one  to  the 
other,  still  have  common  measures,  and  are  therefore 
numerically  equal. 

47.  Similar  Figures  have  equal  angles,  and  the  same 
number  of  sides. 

Polygons  may  be  similar  without  being  equal ;  that  is,  the  angles 
and  the  number  of  sides  may  be  equal,  and  the  length  of  the  sides 
and  the  size  of  the  figures  unequal. 

THE    CIRCLE. 

48.  A  Circle  is  a  plane  figure  bound- 
ed by  one  uniformly  curved  line,  all  of 
the  points  in  which  are  at  the  same 
distance  from  a  certain  point  within, 
called  the  Center. 

49.  The  Circumference  of  a  circle  is 
the  curved  line  that  bounds  it. 


DEFINITIONS.  15 

50.  The  Diameter  of  a  circle  is  a  line  passing  through 
its  center,  and  terminating  at  both  ends  in  the  circum- 
ference. 

51.  The  Radius  of  a  circle  is  a  line  extending  from 
its  center  to  any  point  in  the  circumference.  It  is  one 
half  of  the  diameter.  All  the  diameters  of  a  circle  are 
equal,  as  are  also  all  the  radii. 

52.  An  Arc  of  a  circle  is  any  portion  of  the  circum- 
ference. 

53.  An  angle  having  its  vertex  at  the  center  of  a 
circle  is  measured  by  the  arc  intercepted  by  its  sides. 
Thus,  the  arc  AB  measures  the  angle  AOB;  and  in  gen- 
eral, to  compare  different  angles,  we  have  but  to  compare 
the  arcs,  included  by  their  sides,  of  the  equal  circles 
having  their  centers  at  the  vertices  of  the  angles. 

UNITS  OF  MEASURE. 

54.  The  Numerical  Expression  of  a  Magnitude  is  a  number 
expressing  how  many  times  it  contains  a  magnitude  of  the 
same  kind,  and  of  known  value,  assumed  as  a  unit. 
For  lines,  the  measuring  unit  is  any  straight  line  of  fixed 
value,  as  an  inch,  a  foot,  a  rod,  etc. ;  and  for  surfaces,  the 
measuring  unit  is  a  square  whose  side  may  be  any  linear 
unit,  as  an  inch,  a  foot,  a  mile,  etc.  The  linear  unit 
being  arbitrary,  the  surface  unit  is  equally  so ;  and  its 
selection  is  determined  by  considerations  of  convenience 
and  propriety. 

For  example,  the  parallelogram  ABB  C  is  mea-    c  D 

sured  by  the  number  of  linear  units  in  CB,  mul- 
tiplied by  the  number  of  linear  units  in  A  C  or 
BB  j  the  product  is  the  square  units  in  ABB  C. 
For,  conceive  CB  to  be  composed  of  any  number 
of  equal  parts — say  five — and  each  part  some  unit  of  linear  measure, 
and  AC  composed  of  three  such  units;  from  each  point  of  divi- 
sion on  CD  draw  lines  parallel  *o  A  C,  and  from  each  point  of  divi- 
sion on  AC  draw  lines  parallel  to  CB  or  AB}  then  it  is  as  obvious 


16  GEOMETRY. 

as  an  axiom  that  the  parallelogram  will  contain  5  x  3  =  15  square 
units.  Hence,  to  find  the  areas  of  right-angled  parallelograms,  mul- 
tiply the  base  by  the  altitude. 


EXPLANATION  OF  TERMS. 

55.  An  Axiom  is  a  self-evident  truth,  not  only  too  sim- 
ple to  require,  but  too  simple  to  admit  of,  demonstration. 

56.  A  Proposition  is  something  which  is  either  pro- 
posed to  be  done,  or  to  be  demonstrated,  and  is  either  a 
problem  or  a  theorem. 

57.  A  Problem  is  something  proposed  to  be  done. 

58.  A  Theorem  is  something  proposed  to  be  demon- 
strated. 

59.  A  Hypothesis  is  a  supposition  made  with  a  view  to 
draw  from  it  some  consequence  which  establishes  the 
truth  or  falsehood  of  a  proposition,  or  solves  a  problem. 

60.  A  Lemma  is  something  which  is  premised,  or  demon- 
strated, in  order  to  render  what  follows  more  easy. 

61.  A  Corollary  is  a  consequent  truth  derived  imme- 
diately from  some  preceding  truth  or  demonstration. 

62.  A  Scholium  is  a  remark  or  observation  made  upon 
something  going  before  it. 

63.  A  Postulate  is  a  problem,  the  solution  of  which  is 
self-evident. 

POSTULATES. 

Let  it  be  granted — 

I.  That  a  straight  line  can  be  drawn  from  any  one  point 
to  any  other  point ; 

II.  That  a  straight  line  can  be  produced  to  any  distance, 
or  terminated  at  any  point ; 

III.  That  the  circumference  of  a  circle  can  be  de- 
scribed about  any  center,  at  any  distance  from  that  center. 


DEFINITIONS.  17 

AXIOMS. 

1.  Tilings  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

2.  When  equals  are  added  to  equals  the  wholes  are  equal. 

3.  When  equals  are  taken  from  equals  the  remainders  are 
equal. 

4.  When  equals  are  added  to  unequals  the  wholes  are 
unequal. 

5.  When  equals  are  taken  from  unequals  the  remainders 
are  unequal. 

6.  Things  which  are  double  of  the  same  thing,  or  equal 
things,  are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  Every  whole  is  equal  to  all  its  parts  taken  together. 

10.  Things  which  coincide,  or  fill  the  same  space,  are 
identical,  or  mutually  equal  in  all  their  parts. 

11.  All  right  angles  are  equal  to  one  another. 

12.  A  straight  line  is  the  shortest  distance  between  two 
points. 

13.  Two  straight  lines  cannot  inclose  a  space. 

ABBREVIATIONS. 

The  common  algebraic  signs  are  used  in  this  work, 
and  demonstrations  are  sometimes  made  through  the 
medium  of  equations;  and  it  is  so  necessary  that  the 
student  in  geometry  should  understand  some  of  the  more 
simple  operations  of  algebra,  that  we  assume  that  he  is 
acquainted  with  the  use  of  the  signs.  As  the  terms 
circle,  angle,  triangle,  hypothesis,  axiom,  theorem,  cor- 
ollary, and  definition,  are  constantly  occurring  in  a  course 
of  geometry,  we  shall  abbreviate  them  as  shown  in  the 
following  list : 

2*  B 


18  GEOMETRY. 

Addition  is  expressed  by 4- 

Subtraction         "  "        .         .         .         .         — 

Multiplication    "  "  ....      X 

Equality  and  Equivalency  are  expressed  by  .  = 
Greater  than,  is  expressed  by  .  .  .  .  > 
Less  than,  "  "  < 

Thus :  B  is  greater  than  J.,  is  written    .      B^>A 
B  is  less  than  4,         "      "  .  B<A 

A  circle  is  expressed  by O 

An  angle       "  "  [__ 

A  right  angle  is  expressed  by  .  .  .  E.  [_ 
Degrees,  minutes,  and  seconds,  are  expressed 

by '   .         .         .     °  ' " 

A  triangle  is  expressed  by  ....     A 

The  term  Hypothesis  is  expressed  by     .         .    (Hy.) 
"  Axiom  "  "     .  (Ax.) 

"  Theorem       "  "•  (Th.) 

"  Corollary      «  «     .  (Cor.) 

"  Definition     "  «  (def.) 

"  Perpendicular  is  expressed  by     .  J- 

The  difference  of  two  quantities,  when  it  is 
not  known  which  is  the  greater,  is  ex- 
pressed by  the  symbol  ru 

Thus   ;  the  difference  between  A  and  B  is  written 
A  r^B. 


BOOK   I.  .  19 

BOOK  I. 


OF  STRAIGHT  LINES,  ANGLES,  AND  POLYGONS. 
THEOREM  I. 

When  one  straight  line  meets  another,  not  at  its  extremity, 
the  two  angles  thus  formed  are  two  right  angles,  or  they  are 
together  equal  to  two  right  angles. 

Let  AB  meet  CD,  and  if  AB  is  perpen-  ¥  ,A 

dicular  to  CD,  it  does  not  incline  to  either 
extremity  of  CD.     In  that  case,  the  angle 
ABD  is  equal  to  the  angle  ABC,  and  is  c" 
a  right  angle,  by  Definition  15. 

But  if  these  angles  are  unequal,  we  are  to  show  that 
their  sum  is  equal  to  two  right  angles.  Conceive  the 
dotted  line  BE  to  be  drawn  from  the  point  B,  so  as  not  to 
incline  to  either  side  of  CD ;  then,  by  Def.  15,  the  angles 
<  CBE  and  JEJBD  are  right  angles ;  but  the  angles  CB A 
and  ABD  make  the  same  sum,  or  fill  the  same  angular 
space,  as  the  two  angles  CBE  and  EBD,  and  are,  con- 
sequently, equal  to  two  right  angles.  Hence  the  theorem ; 
when  one  straight  line  meets  another,  not  at  its  extremity,  the 
sum  of  the  two  angles  is  equal  to  two  right  angles. 

Cor.  Hence,  the  two  angles  ABC  and  ABD  are  supple- 
mentary to  each  other,  (Def.  21). 

THEOREM    II. 

From  any  point  in  a  straight  line,  not  at  its  extremity,  the 
sum  of  all  the  angles  that  can  be  formed  on  the  same  side  of 
the  line  is  equal  to  two  right  angles. 

Let  CD  be  any  line,  and  B  any  point     w  f 

in  it. 

We  are  to  show  that  the  sum  of  all  the 
angles  which  can  be  formed  at  B,  on  one  c 
side  of  CD,  will  be  equal  to  two  right  angles. 


20 


GEOMETRY. 


By  Th.  1,  any  two  supplementary  angles,  as  ABB, 
ABO,  are  together  equal  to  two  right  angles.  And  since 
the  angular  space  about  the  point  B  is  neither  increased 
nor  diminished  by  the  number  of  lines  drawn  from  that 
point,  the  sum  of  all  the  angles  DBA,  ABU,  EBH, 
HBO,  fills  the  same  spaces  as  any  two  angles  HBB, 
HBO.  Hence  the  theorem  ;  from  any  point  in  a  line,  the 
sum  of  all  the  angles  that  can  be  formed  on  the  same  side  of 
the  line  is  equal  to  two  right  angles. 

Oor.  1.  And,  as  the  sum  of  all  the  angles  that  can  be 
formed  on  the  other  side  of  the  line,  OB,  is  also  equal  to 
two  right  angles ;  therefore,  all  the  angles  that  can  be 
formed  quite  round  a  point,  B,  by  any  number  of  lines,  are 
together  equal  to  four  right  angles. 

Oor.  2.  Hence,  also,  the  whole  circum- 
ference of  a  circle,  being  the  sum  of  the 
measures  of  all  the  angles  that  can  be 
made  about  the  center  F,  (Def.  53),  is  the 
measure  of  four  right  angles;  conse- 
quently, a  semicircle,  or  180°,  is  the  mea- 
sure of  two  right  angles ;  and  a  quadrant,  or  90°,  is  the 
measure  of  one  right  angle. 


THEOREM   III. 

If  one  straight  line  meets  two  other  straight  lines  at  a 
common  point,  forming  two  angles,  which  together  are  equal 
to  two  right  angles,  the  two  straight  lines  are  one  and  the 
same  line. 

Let  the  line  AB  meet  the 
lines  BD  and  BE  at  the  com- 
mon point  B,  making  the  sum 
of  the  two  angles  ABB,  ABE, 
equal  to  two  right  angles ;  we 
are  to  prove  that  DB  and  BE 
are  one  straight  line. 


BOOK    I.  21 

If  BB  and  BE  are  not  in  the  same  line,  produce  BB 
to  0,  thus  forming  one  line,  BBO. 

Now  by  Th.  1,  ABB  +  ABO  must  be  equal  to  two 
right  angles.  But  by  hypothesis,  ABB  +  ABE  is  equal 
to  two  right  angles. 

Therefore,  ABB  +  ABO  is  equal  to  ABB  +  ABE, 
(Ax.  1).  From  each  of  these  equals  take  away  the  com- 
mon angle  ABB,  and  the  angle  ABO  will  be  equal  to 
ABE,  (Ax.  5).  That  is,  the  line  BE  must  coincide  with 
BO,  and  they  will  be  in  fact  one  and  the  same  line,  and 
they  cannot  be  separated  as  is  represented  in  the  figure. 

Hence  the  theorem ;  if  one  line  meets  two  other  lines  at  a 
common  point,  forming  two  angles  which  together  are  equal 
to  two  right  angles,  the  two  lines  are  one  and  the  same  line. 

THEOKEM    IV. 

If  two  straight  lines  intersect  each  other,  the  opposite  or 
vertical  angles  must  be  equal. 

If  AB  and  OB  intersect  each 
other  at  E,  we  are  to  demonstrate 
that  the  angle  AEO  is  equal  to 
the  vertical  angle  DEB ;  and  the 
angle  AEB,  to  the  vertical  angle 
OEB. 

As  AB  is  one  line  met  by  BE,  another  line,  the  two 
angles  AEB  and  BEB,  on  the  same  side  of  AB,  are  equal 
to  two  right  angles,  (Th.  1).  Also,  because  OB  is  a  right 
line,  and  AE  meets  it,  the  two  angles  AEO  and  AEB 
are  together  equal  to  two  right  angles. 

Therefore,  AEB  +  BEB  ft-  AEO  +  AEB.     (Ax.  1.) 

If  from  these  equals  we  take  away  the  common  angle 
AEB,  the  remaining  angle  BEB  must  be  equal  to  the 
remaining  angle  AEO,  (Ax.  3).  In  like  manner,  we  can 
prove  that  AEB  is  equal  to  OEB.  Hence  the  theorem ; 
if  the  two  lines  intersect  each  other,  the  vertical  angles  must 
be  equal. 


22        .  GEOMETRY. 

Second  Demonstration, 

By  Def.  11,  the  angle  DEB  is  the  difference  in  the 
direction  of  the  lines  ED  and  EB;  and  the  angle  AEQ 
is  the  difference  in  the  direction  of  the  lines  EC  and  EA. 

But  ED  is  opposite  in  direction  to  EQ;  and  EB  is 
opposite  in  direction  to  EA. 

Hence,  the  difference  in  the  direction  of  ED  and  EB 
is  the  same  as  that  of  EQ  and  EA,  as  is  ohvious  by  in- 
spection. 

Therefore,  the  angle  DEB  is  equal  to  its  opposite  AEC. 

In  like  manner,  we  may  prove  AED  =  CEB. 

Hence  the  theorem ;  if  two  lines  intersect  each  other,  the 
vertical  angles  must  be  equal. 

THEOREM    V. 

If  a  straight  line  intersects  two  parallel  lines,  the  sum  of 
the  two  interior  angles  on  the  same  side  of  the  intersecting 
line  is  equal  to  two  right  angles. 

[Note.  —  By  interior  angles,  we  mean  angles  which  lie  between  the 
parallels ;  the  exterior  angles  are  those  not  between  the  parallels.] 

Let  the  parallel  lines  AB 
and  CD  intersect  EF;  then 
we  are  to  demonstrate  that 

the  angles  BGH  +  GHD  =        

2  R,  L  

Because  GB  and  ED  are       c  /H  d 

parallel,  they  are  equally  in-  jf 

clined  to  the  line  EF,  or  have 

the  same  difference  of  direction  from  that  line.  There- 
fore, [_FGB  =  L  &HD.  To  each  of  these  equals  add 
the  [_BGH,  and  we  have  FGB +BGH=GHD+BGH. 

But  by  Th.  1,  the  first  member  of  this  equation  is  equal 
to  two  right  angles ;  and  the  second  member  is  the  sum 
of  the  two  angles  between  the  parallels.  Hence  the  theo- 
rem ;  if  a  line  intersects  two  parallel  lines,  the  sum  of  the  two 
interior  angles  on  the  same  side  of  the  intersecting  line  must 
be  equal  to  two  right  angles. 


BOOK    I.  23 

Scholium.  —  As  AB  and  CD  are  parallel  lines,  and  EF  is  a  lino 
intersecting  them,  AB  and  EF  must  make  equal  angles  to  those  made 
by  CD  and  EF.  That  is,  the  angles  about  the  point  G  must  be  equal 
to  the  corresponding  angles  about  the  point  H. 

THEOREM    VI. 

If  a  line  intersects  two  parallel  lines,  the  alternate  interior 
angles  are  equal. 

Let  AB  and  CD  be  paral- 
lels, intersected  by  EF  at  R 
and  G.    Then  we  are  to  prove 

that  the  angle  A  GH  is  equal ^~y — 

to  the  alternate  angle  GHD,  / 

and  CHG  -  HGB.  c /H  f> 

By  Th.  5,  [_BGH  +  ]_  / 

GrHD  —  two  right  angles.  Al- 
so, by  Th.  1,  [_AGH  +  [_BGH  =  two  right  angles. 
From  these  equals  take  away  the  common  angle  BGH, 
and  L  CHD  will  be  left,  equal  to  \_AGH,  (Ax.  3).  In 
like  manner,  we  can  prove  that  the  angle  CHG  is  equal 
to  the  angle  HGB.  Hence  the  theorem ;  if  a  line  intersects 
two  parallel  lines,  the  alternate  interior  angles  are  equal. 

Cor.  1.  Since        |__  A  GH  =  [__  FGB, 

and  [__AGH=l_GHD; 

Therefore,  [_  FGB  =  \__  GHD  (Ax.  1). 

Also,  [__AGF  +  l_AGH=2H.  |___,  (Th.  1), 

and  L  CHG  +  [_AGH  =  2  K.  [_,  (Th.  5); 

Therefore, 

[_AGF+]_AGH  =  ]_CHG  +  [__AGH,  (Ax.l); 

and  L.AGF  =  [_  CHG,  (Ax.  3). 

That  is,  the  exterior  angle  is  equal  to  the  interior  opposite 
angle  on  the  same  side  of  the  intersecting  line. 

Cor.  2.  Since      [__AGH  =  [__FGB, 

and  \_AGH=[_CHE; 

Therefore,  l__FGB  =  L  CHE. 

In  the  same  manner  it  may  be  shown  that 
[__AGF  =  [_EHD. 

Hence,  the  alternate  exterior  angles  are  equal. 


24  GEOMETKY. 

THEOREM    VII. 

If  a  line  intersects  two  other  lines,  making  the  sum  of  the 
two  interior  angles  on  the  same  side  of  the  intersecting  line 
equal  to  two  right  angles,  the  two  straight  lines  are  parallel. 

Let  the  line  FF  intersect 
the  lines  AB  and  CB,  making 
the  two  angles  BaR  +  GHD  A 
=  to  two  right  angles ;  then 
we  are  to  demonstrate  that 
AB  and  OB  are  parallel.  C  /H  ~~b 

As  EF  is  a  right  line  and  B' 

BCr  meets  it,  the  two  angles 

FGrB  and  BGrlT  are  together  equal  to  two  right  angles, 
(Th.  1).  But  by  hypothesis,  the  angles,  j?##and  aHB, 
are  together  equal  to  two  right  angles.  From  these  two 
equals  take  away  the  common  angle  BGrH,  and  the  re- 
maining angles  FGrB  and  CrffB  must  be  equal,  (Ax.  3). 
'Now,  because  GfB  and  SB  make  equal  angles  with  the 
same  line  EF,  they  must  extend  in  the  same  direction ;  and 
lines  having  the  same  direction  are  parallel,  (Def.  13). 
Hence  the  theorem ;  if  a  line  intersects  two  other  lines,  making 
the  sum  of  the  two  interior  angles  on  the  same  side  of  the  in- 
tersecting line  equal  to  two  right  angles,  the  two  lines  must  be 
parallel. 

Cor.  1,  Ifa  line  intersects  two  other  lines,  making  the 
alternate  interior  angles  equal,  the  two  lines  intersected 
must  be  parallel. 

Suppose  the  L  -4##  =  L  ff^Z>-  Adding  \__HGB 
to  each,  we  have 

[__AGH  +  L HGLB  =  L  aEI)  +  L.B&B- 

but  the  first  member  of  this  equation,  that  is,  [_AGR-\- 
|__  HGrB,  is  equal  to  two  right  angles ;  hence  the  second 
member  is  also  equal  to  the  same ;  and  by  the  theorem, 
the  lines  AB  and  CD  are  parallel. 

Cor.  2.  If  a  line  intersects  two  other  lines,  making  the 


BOOK    I.  25 

opposite  exterior  and  interior  angles  equal,  the  two  lines 
intersected  must  be  parallel. 

Suppose  the  [__ FGB  =  |_  &#B.    Adding  the  [__EaB 
to  each,  we  have 

L fgb  +  \_hgb  =  i  ®HI>  +  EaB- 

But  the  first  member  of  this  equation  is  equal  to  two 
right  angles  ;  hence  the  second  member  is  also  equal  to 
two  right  angles ;  and  by  the  theorem,  the  lines  AB  and, 
CD  are  parallel. 

Cor.  3.  If  a  line  intersects  two  other  lines,  making  the 
alternate  exterior  angles  equal,  the  lines  must  be  parallel, 

Suppose     [_BGF=\_CHE,  and  [_AGF  =  [_DHE, 

ByTh.4,     \_BGF=[_AGH,^di[_OHE^l_DHG. 

And  since  [_BGF =  [_OHE,         [_AGH=\_DHG. 

That  is,  the  alternate  interior  angles  are  equal;  and 
hence  (by  Cor.  1)  the  two  lines  are  parallel. 

THEOREM    VIII. 

If  two  angles  have  their  sides  parallel,  the  two  angles  will 
be  either  equal  or  supplementary. 

Let  A  0  be  parallel  to  BD,  and  AH 
parallel  to  BF  or  to  BG.    Then  we  are 
to  prove  that  the  angle  DBF  is  equal 
to  the  angle  CAH,  and  that  the  angle 
DBG  is  supplementary  to  the  angle  A. 
The  angle  OAH  is  formed  by  the  differ- 
ence in  the  direction  of  A  C  and  AH;  and 
the  angle  DBF  is  formed  by  the  differ- 
ence in  the  direction  of  BD  and  BF. 
But  AC  and  AH  have  the  same  direc- 
tions as  BD  and  BF,  because  they  are  respectively  paral- 
lel.   Therefore,  by  Def.  11,  L  CAH=  [_DBF.     But  the 
line  BG  has  the  same  direction  as  BF,  and  the  angle 
DBG  is  supplementary  to  DBF.     Hence  the  theorem; 
angles  whose  sides  are  parallel,  form  either  equal  or  supple- 
mentary  angles. 
3 


26  GEOMETRY. 

THEOREM    IX. 

The  opposite  angles  of  any  parallelogram  are  equal. 

Let  AEBGi  be  a  parallel- 
ogram.     Then   we    are   to  \ 

prove  that  the  angle  GBE     G _\B_ 

is  equal  to  its  opposite  angle 
A. 

*    Produce  EB  to  D,  and  GB 
to  F;  then,  since  BJ)  is  par- 
allel to  A  G,  and  BF  to  AE,  the  angle  DBF  is  equal  to 
the  angle  A,  (Th.  8). 

But  the  angles  GBE  and  DBF,  being  vertical,  are 
equal,  (Th.  4).  Therefore,  the  opposite  angles  QBE  and 
A,  of  the  parallelogram  AEBG,  are  equal. 

In  like  manner,  we  can  prove  the  angle  E  equal  to 
the  angle  G.  Hence  the  theorem ;  the  opposite  angles  of 
any  parallelogram  are  equal, 

THEOREM    X. 

The  sum  of  the  angles  of  any  parallelogram  is  equal  to 
four  right  angles. 

Let  ABCB  be  a  parallelo- 
gram. We  are  to  prove  that 
the  sum  of  the  angles  A,  B,  0 
and  B,  is  equal  to  four  right 
angles,  or  to  360°. 

Because  AB  and  BC  are  parallel  lines,  and  AB  inter- 
sects them,  the  two  interior  angles  A  and  B  are  together 
equal  to  two  right  angles,  (Th.  5).  And  because  CD  in- 
tersects the  same  parallels,  the  two  interior  angles  C  and 
D  are  also  together  equal  to  two  right  angles.  By  addi- 
tion, we  have  the  sum  of  the  four  interior  angles  of  the 
parallelogram  ABCB,  equal  to  four  right  angles.  Hence 
the  theorem ;  the  sum  of  the  angles  of  any  parallelogram  is 
equal  to  four  right  angles. 


BOOK    I. 


THEOREM    XI, 


The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two 
right  angles. 

Let  A B  0  be  a  triangle, 
and  through  its  vertex  0 
draw  a  line  parallel  to  the 
base  AB,  and  produce 
the  sides  AC  and  BO. 
Then  the  angles  A  and 
a,  being  exterior  and  in- 
terior opposite  angles  on 
the  same  side  of  two  parallels,  are  equal,  (Th.  6,  Cor.  1). 
For  like  reasons,  \__B  =  [__£.  And  the  angles  0  and  c, 
being  vertical  angles,  are  also  equal,  (Th.  4).  Therefore, 
the  angles  A,  B,  C  are  equal  to  the  angles  a,  b,  c  respect- 
ively. But  the  angles  around  the  point  (7,  on  the  upper 
side  of  the  parallel  CD,  are  equal  to  two  right  angles, 
(by  Th.  1).  Hence  the  theorem;  the  sum  of  the  three 
angles,  etc. 

Second  Demonstration. 

hetAEBG  be  a  parallelogram. 
Draw  the  diagonal  GE;  then  the 
parallelogram  is  divided  into  two 
triangles,  and  the  opposite  angles  E  R 

€r  and  E  are  mutually  divided  by  the  diagonal  GrE. 

Because  GrB  and  AE  are  parallel,  the  alternate  interior 
angles  BGrE  and  GrEA  are  equal,  (Th.  6).  Designate 
each  of  these  by  b. 

In  like  manner,  because  EB  and  A  G-  are  parallel,  the 
alternate  interior  angles,  BEG-  and  EGA,  are  equal. 
Designate  each  of  these  by  a. 

Now  we  are  to  prove  that  the  three  angles  B>  b,  and  a, 
and  also  that  the  three  angles  A,  a,  and  b,  are  equal  to 
two  right  angles. 


28  GEOMETRY. 

Because  A  and  B  are  opposite  angles  of  a  parallelo- 
gram, they  are  equal,  (Th.    9),  and  [_A  +  [_B  =  2  \__A. 

And  all  the  interior  angles  of  the  parallelogram  are 
equal  to  four  right  angles,  (Th.  10). 

Therefore,  2 A  +  2a  -f  26  =  4  right  angles. 

Dividing  by  2,  and  A  4-    a  +    b  =  2  " 

That  is,  all  the  angles  of  the  triangle  AGE  are  together 
equal  to  two  right  angles. 

Hence  the  theorem ;  the  sum  of  the  three  angles,  etc. 

Scholium. — Any  triangle,  as  AGE,  may  be  conceived  to  be  part  of 
a  parallelogram.  For,  let  A  GE  be  drawn  independently  of  the  paral- 
lelogram ;  then  draw  EB  from  the  point  E  parallel  to  A  G,  and  through 
the  point  G  draw  GB  parallel  to  AE,  and  a  parallelogram  will  be 
formed  embracing  the  triangle ;  and  thus  the  sum  of  the  three  angles 
of  any  triangle  is  proved  equal  to  two  right  angles. 

This  truth  is  so  fundamental,  important,  and  practical, 
as  to  require  special  attention ;  we  therefore  give  a 

Third  Demonstration. 

Let  ABO  be  a  triangle.  Then 
we  are  to  show  that  the  angles  A, 
0,  and  ABC,  are  together  equal 
to  two  right  angles. 

Let  AB  be  produced  to  D,  and 
from  B  draw  BE  parallel  to  AC. 

Then,  EBB  and  OAB  being  exterior  and  interior  op- 
posite angles  on  the  same  side  of  the  line  AB,  are  equal, 
(Th.  6,  Cor.  1).  Also,  QBE  and  ACB,  being  alternate 
angles,  are  equal,  (Th.  6). 

By  addition,  observing  that  [__  QBE,  added  to  [_EBB, 
must  make  [_  CBD,  we  have 

[_CBB  =  l_A  +  l_a  (1.) 

To  each  of  these  equals  add  the  angle  CBA,  and  we 
shall  have 

[_CBA  +  [_CBD=  L_^  +  l_C+l_CBA. 
But  (by  Th.  1),  the  sum  of  the  first  two  is  equal  to  two 


BOOK   I.  29 


right  angles;  therefore,  the  three  angles,  A,  0,  and  OB  A, 
\re  together  equal  to  two  right  angles. 
Hence  the  theorem ;  the  sum  of  the  three  angles,  etc. 


THEOREM    XII. 

If  any  side  of  a  triangle  is  'produced,  the  exterior  angle  is 
equal  to  the  sum  of  the  two  interior  opposite  angles. 

Let  ABO  be  a  triangle.  Pro- 
duce AB  to  D;  and  we  are  to 
prove  that  the  angle  OBI)  is  equal 
to  the  sum  of  the  two  angles  A 
and  O. 

We  establish  this  theorem  by  a 
course  of  reasoning  in  all  respects  the  same  as  that  by 
which  we  obtained  Eq.  (1.),  third  demonstration,  (Th.  11). 

Oor.  1.  Since  the  exterior  angle  of  any  triangle  is  equal 
to  the  sum  of  the  two  interior  opposite  angles,  therefore 
it  is  greater  than  either  one  of  them. 

Oor.  2.  If  two  angles  in  one  triangle  be  equal  to  two 
angles  in  another  triangle,  the  third  angles  will  also  be 
equal,  each  to  each,  (Ax.  3) ;  that  is,  the  two  triangles 
will  be  mutually  equiangular. 

Oor.  3.  If  one  angle  in  a  triangle  be  equal  to  one  angle 
in  another,  the  sum  of  the  remaining  angles  in  the  one 
will  also  be  equal  to  the  sum  of  the  remaining  angles  in 
the  other,  (Ax.  3). 

Oor.  4.  If  one  angle  of  a  triangle  be  a  right  angle,  the 
sum  of  the  other  two  will  be  equal  to  a  right  angle, 
and  each  of  them  singly  will  be  acute,  or  less  than  a  right 
angle. 

Oor.  5.  The  two  smaller  angles  of  every  triangle  are 
acute,  or  each  is  less  than  a  right  angle. 

Oor.  6.  All  the  angles  of  a  triangle  may  be  acute,  but 
no  triangle  can  have  more  than  one  right  or  one  obtuse 
angle. 


30  GEOMETRY. 

THEOREM    XIII. 

In  any  quadrilateral,  the  sum  of  the  four  interior  angles  is 
equal  to  four  right  angles. 

Let  ABCD  be  a  quadrilateral;  then  we 
are  to  prove  that  the  sum  of  the  four  in- 
terior angles,  that  is  A  -f  B  +  0  +  D,  is 
equal  to  four  right  angles. 

Draw  the  diagonal  AC,  dividing  the 
quadrilateral  into  two  triangles,  ABO, 
ABO.  Now,  since  the  sum  of  the  three  angles  of  each 
of  these  triangles  is  equal  to  two  right  angles,  (Th.  11), 
it  follows  that  the  sum  of  all  the  angles  of  both  triangles 
which  make  up  the  four  angles  of  the  quadrilateral,  must 
be  equal  to  four  right  angles,  (Ax.  2). 

Hence  the  theorem ;  in  any  quadrilateral,  etc. 

Cor.  1.  Hence,  if  three  of  the  angles  of  a  quadrilateral 
are  right  angles,  the  fourth  will  also  be  a  right  angle. 

Oor.  2.  If  the  sum  of  two  of  the  four  angles  be  equal 
to  two  right  angles,  the  sum  of  the  remaining  two  will 
also  be  equal  to  two  right  angles.  And,  if  the  sum  of 
either  two  of  the  angles  be  less  than  two  right  angles, 
the  sum  of  the  other  two  angles  will  be  greater  than  two 
right  angles. 

THEOREM   XIV. 

In  any  polygon,  the  sum  of  all  the  interior  angles  is  equal 
to  twice  as  many  right  angles,  less  four,  as  the  figure  has  sides. 

Let  ABODE  be  any  polygon ; 
we  are  to  prove  that  the  sum  of 
all  its  interior  angles,  A  +  B  -f  0 
-f-  D  +  E,  is  equal  to  twice  as 
many  right  angles,  less  four,  as 
the  figure  has  sides. 

From  any  point,  p,  within  the 
figure,  draw  lines  pA,  pB,  pO,  etc.,  to  all  the  angles, 


BOOK    I.  31 

thus  dividing  the  polygon  into  as  many  triangles  as  it 
has  sides.  Now,  the  sum  of  the  three  angles  of  each  of 
these  triangles  is  equal  to  two  right  angles,  (Th.  11) ;  and 
the  sum  of  the  angles  of  all  the  triangles  must  be  equal 
to  twice  as  many  right  angles  as  the  figure  has  sides. 
But  the  sum  of  these  angles  contains  the  sum  of  four 
right  angles  about  the  point  p ;  taking  these  away,  and 
the  remainder  is  the  sum  of  the  interior  angles  of  the 
figure.  Therefore,  the  sum  must  be  equal  to  twice  as 
many  right  angles,  less  four,  as  the  figure  has  sides. 

Hence  the  theorem ;  in  any  polygon,  etc. 

From  this  Theorem  is  derived  the  rule  for  finding  the 
sum  of  the  interior  angles  of  any  right-lined  figure : 

Subtract  2  from  the  number  of  sides,  and  multiply  the  re- 
mainder by  2  ;  the  product  will  be  the  number  of  right  angles. 

Thus,  if  the  number  of  sides  be  represented  by  S,  the 
number  of  right  angles  will  be  represented  by  (2S  —  4). 

The  Theorem  is  not  varied  in 
case  of  a  re-entrant  angle,  as  rep-                   ^1 
resented  at  d,  in  the  figure  ABC-          ^^ 
DEF.  ^-------JA 

Draw  lines   from  the   angle  d        \       /'   \        / 
to   the    several    opposite    angles,         \  /  \  / 

making  as  many  triangles  as  the 
figure  has  sides,  less  two,  and  the 

sum  of  the  three  angles  of  each  triangle  equals  two  right 
angles. 

THEOREM   XV. 

From  any  point  without  a  straight  line,  but  one  perpendic- 
ular can  be  drawn  to  that  line. 

From  the  point  A  let  us  suppose  A 
it  possible  that  two  perpendiculars, 
A B  and  A  C,  can  be  drawn.  Now,  be- 
cause AB  is  a  supposed  perpendicu- 
lar, the  angle  ABC  is  a  right  angle ;  .    


and  because  A  C  is  a  supposed  per-  B  c 


32  GEOMETKY. 

pendicular,  the  angle  A  CB  is  also  a  right  angle;  and  if 
two  angles  of  the  triangle  ABC  are  together  equal  to  two 
right  angles,  the  third  angle,  BAG,  must  be  infinitely 
small,  or  zero ;  but  this  is  impossible,  for  it  requires  the 
sum  of  the  three  angles  of  a  triangle  to  make  two  right 
angles,  (Th.  11).  Therefore,  the  lines  AB  and  A 0  must 
be  identical,  or  but  one  perpendicular. 

Hence  the  theorem  ;  from  any  point  without  a  straight 
line,  etc. 

Cor.  At  a  given  point  in  a  straight  line  but  one  per- 
pendicular can  be  erected  to  that  line ;  for,  if  there  could 
be  two  perpendiculars,  we  should- have  unequal  right 
angles,  which  is  impossible. 

THEOREM    XYI. 

Two  triangles  which  have  two  sides  and  the  included  angle 
in  the  one,  equal  to  two  sides  and  the  included  angle  in  the 
other,  each  to  each,  are  equal  in  all  respects. 

In  the  two  A's,  ABC  and  BEF, 
on  the  supposition  that  AB  =  BE, 
AC=BF,  and  [_A  =  [_B,  we 
are  to  prove  that  BC  must  =  EF, 
the  [__B  =  L.2J,  and  the  [_C  = 
l_F. 

Conceive  the  a  ABC  cut  out  of  the  paper,  taken  up, 
and  placed  on  the  A  BEF  in  such  a  manner  that  the 
point  A  shall  fall  on  the  point  B,  and  the  line  AB  on 
the  line  BE;  then  the  point  B  will  fall  on  the  point  E, 
because  the  lines  are  equal.  Now,  as  the  [__A  =  [__B, 
the  line  A  C  must  take  the  same  direction  as  BF,  and  fall 
on  BF;  and  as  AC  =  BF,  the  point  C  will  fall  on  F.  B 
being  on  E  and  C  on  F,  BC  must  be  exactly  on  EF. 
(otherwise,  two  straight  lines  would  enclose  a  space,  Ax. 
13),  and  BC '=  EF,  and  the  two  magnitudes  exactly  fill 
the  same  space.  Therefore,  BC  =  EF,  [__B  =  [_E, 
L  (7=  [_F,  and  the  two  A's  are  equal,  (Ax.  9). 

Hence  the  theorem ;  two  triangles  which  have  two  sides,  etc. 


BOOK    I. 


33 


THEOREM   XVII. 

When  two  triangles  have  a  side  and  two  adjacent  angles  in 
the  one,  equal  to  a  side  and  two  adjacent  angles  in  the  other, 
each  to  each,  the  two  triangles  are  equal  in  all  respects. 

In  two  a's,  as  ABO  and 
DEF,  on  the  supposition 
that  BO  =  EF,[_B=[_E, 
and  [v  0  —  [__ F,  we  are  to 
prove  that  AB  m  BE,  AG 
=  DF,  andL^-  -  L-^- 

Conceive  the  A  ABO  taken  np  and  placed  on  the  A 
BEF,  so  that  the  side  BO  shall  exactly  coincide  with  its 
equal  side  EF;  now,  because  the  angle  B  is  equal  to  the 
angle  E,  the  line  BA  will  take  the  direction  of  ED,  and 
will  fall  exactly  upon  it ;  and  because  the  angle  0  is  equal 
to  the  angle  F,  the  line  OA  will  take  the  direction  of 
FD,  and  fall  exactly  upon  it ;  and  the  two  lines  BA  and 
OA,  exactly  coinciding  with  the  two  lines  ED  and  FD, 
the  point  A  will  fall  on  D,  and  the  two  magnitudes  will 
exactly  fill  the  same  space ;  therefore,  by  Ax.  10,  they  are 
equal,  and  AB  =  DE,  AO=DF,  and  the  \_A  =  [_D. 

Hence  the  theorem ;  when  two  triangles  have  a  side  and 
two  adjacent  angles  in  the  one,  equal  to,  etc. 


THEOREM    XVIII. 

If  two  sides  of  a  triangle  are  equal,  the  angles  opposite 
these  sides  are  also  equal. 

Let  ABO  be  a  triangle;  and  on 
the  supposition  that  AO  =  BO,  we 
are  to  prove  that  the  [__  -4= the  [_B. 

Conceive  the  angle  0  divided  into 
two  equal  angles  by  the  line  OD; 
then  we  have  two  A's,  ADO  and 
BDO,  which  have  the  two  sides,  AO 
and  OD  of  the  one,  equal  to  the  two 
sides,  OB  and  OD  of  the  other ;  and 


to 


34  GEOMETRY. 

the  included  angle  ACD,  of  the  one,  equal  to  the  in- 
cluded angle  BCD  of  the  other:  therefore,  (Th.  16),  AD 
m  BD,  and  the  angle  A,  opposite  to  CD  of  the  one  tri- 
angle, is  equal  to  the  angle  B,  opposite  to  CD  of  the 
other  triangle ;  that  is,  [_A=  [_B. 

Hence  the  theorem ;  if  two  sides  of  a  triangle  are  equal, 
the  angles,  etc. 

Cor.  1.  Conversely :  if  two  angles  of  a  triangle  are  equal, 
the  sides  opposite  to  them  are  equal,  and  the  triangle  is 
isosceles. 

For,  if  A  C  is  not  equal  to  BC,  suppose  BC  to  be  the 
greater,  and  make  BE  =  AE;  then  will  A  AEB  be  isos- 
celes, and  [_EAB  =  \_EBA  ;  hence  [__EAB  =  [_  CAB, 
or  a  part  is  equal  to  the  whole,  which  is  absurd ;  therefore, 
CB  cannot  be  greater  than  AC,  that  is,  neither  of  the 
sides  AC,  BC,  can  be  greater  than  the  other,  and  conse- 
quently they  are  equal. 

Cor.  2.  As  the  two  triangles,  ACD  and  BCD,  are  in  all 
respects  equal,  the  line  which  bisects  the  angle  included 
between  the  equal  sides  of  an  isosceles  A  also  bisects  the 
base,  and  is  perpendicular  to  the  base. 

Scholium  1. — If  in  the  perpendicular  DC,  any  other  point  than  C 
be  taken,  and  lines  be  drawn  to  the  extremities  A  and  B,  such  lines 
will  be  equal,  as  is  evident  from  Th.  16 ;  hence,  we  may  announce 
this  truth :  Any  point  in  a  perpendicular  drawn  from  the  middle  of  a 
line,  is  at  equal  distances  from  the  two  extremities  of  the  line. 

Scholium  2.  —  Since  two  points  determine  the  position  of  a  line,  it 
follows,  that  the  line  which  connects  two  points  equally  distant  from  the 
extremities  of  a  given  line,  is  perpendicular  to  this  line  at  its  middle 
point. 

THEOREM    XIX. 

The  greater  side  of  every  triangle  has  the  greater  angle 
opposite  to  it. 

Let  ABC  be  a  A  ;  and  on  the  supposition  that  A  C  is 
greater  than  AB,  we  are  to  prove  that  the  angle  ABCia 


BOOK    I.  35 

greater  than  the  [_  0.     From  AO,  the 
greater  of  the  two  sides,  take  AB,  equal  ^ 

to  the  less  side  AB,  and  draw  BB,  thus  /\ 

making  two  triangles  of  the  original  tri-  /     \ 

angle.     As  AB  =  AID,  the  [_ABB  =        /       \ 
the  [_  ABB,  (Th.18).  B\T    \ 

But  the  L  ABB  is  the  exterior  angle  \\ 

of  the  A  BBC,  and  is  therefore  greater  C 

than  O,  (Th.  12);  that  is,  the  [__ABB 
is  greater  than  the  angle  0.     Much  more,  then,  is  the 
angle  AB 0  greater  than  the  angle  0. 

Hence  the  theorem ;  the  greater  side  of  every  triangle,  etc. 

Cor.  Conversely:  the  greater  angle  of  any  triangle  has 
the  greater  side  opposite  to  it. 

In  the  triangle  ABO,  let  the  angle  B  be  greater  than 
the  angle  A ;  then  is  the  side  A  0  greater  than  the  side 
BO. 

Tor,  if  BO  —  AO,  the  angle  A  must  be  equal  to  the 
angle  B,  (Th.  18),  which  is  contrary  to  the  hypothesis ; 
and  if  BC^>AC,  the  angle  A  must  be  greater  than  the 
angle  B,  by  what  is  above  proved,  which  is  also  contrary 
to  the  hypothesis ;  hence  BO  can  be  neither  equal  to,  nor 
greater,  than  AO;  it  is  therefore  less  than  AO. 

THEOREM   XX. 

The  difference  between  any  two  sides  of  a  triangle  is  less 

than  the  third  side. 

A 

Let  JLJ? (7 be  a  A,  in  which  JL<7 is  greater  \ 

than  AB;  then  we  are  to  prove  that  AO  I  \ 

— AB  is  less  than  BO.  I     \ 

On  AO,  the  greater  of  the  two  sides,  / ,M) 

lay  off  AB  equal  to  AB.  ^>sJ  \ 

Now,  as  a  straight  line  is  the  shortest  ^^^ 

distance  between  two  points,  we  have  c 

AB  +  BO>AO.  (1) 


GEOMETRY. 


From  these  unequals  subtract  the  equals  AB  —  AD, 
and  we  have  BO  >  AC—  AB.     (Ax.  5). 

Hence  the  theorem ;  the  difference  between  any  two  sides 
of  a  triangle,  etc. 


THEOREM    XXI. 

If  two  triangles  have  the  three  sides  of  the  one  equal  to 
the  three  sides  of  the  other,  each  to  each,  the  two  triangles  are 
equil,  and  the  equal  angles  are  opposite  the  equal  sides. 

In  two  triangles,  as  ABC  and  ABB,  on  the  supposition 
that  the  side  AB  of  the  one  =  the  side  AB  of  the  other, 
AC— AD,  and  BC=BD,  we  are  to  demonstrate  that 
\_ACB=l_ADB,  \_BAC  = 
[_BAD,  and  \__ABC=  \__ABD. 

Conceive  the  two  triangles  to 
be  joined  together  by  their  long- 
est equal  sides,  and  draw  the 
line  CD. 

Then,  in  the  triangle  A  CD, 
because   A  C  is    equal   to  AD, 

the  angle  ACD  is  equal  to  the  angle  ADC,  (Th.  18).  In 
like,  manner,  in  the  triangle  BCD,  because  BC  is  equal 
to  BD,  the  angle  BCD  is  equal  to  the  angle  BDC.  Now, 
the  angle  ACD  being  equal  to  the  angle  ADC,  and  the 
angle  BCD  to  the  angle  BDC,  [_ACD  +  [__BCD=  [_ 
ADC  +[_BDC,  (Ax.  2) ;  that  is,  the  whole  angle  A CB  is 
equal  to  the  whole  angle  ADB. 

Since  the  two  sides  AC  and  CB  are  equal  to  the  two  sides 
AD  and  DB,  each  to  each,  and  their  included  angles  A  CB, 
ADB,  are  also  equal,  the  two  triangles  ABC,  ABD,  are 
equal,  (Th.  16),  and  have  their  other  angles  equal ;  that 
is,  \_BAC=  \_BAD,  and  [_ABC=  [_ABD. 

Hence  the  theorem ;  if  two  triangles  have  the  three  side$ 
of  the  one,  etc. 


BOOK    I. 


THEOREM   XXII. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two 
sides  of  the  other,  each  to  each,  and  the  included  angles  un- 
equal, the  third  sides  will  be  unequal,  and  the  greater  third 
side  will  belong  to  the  triangle  which  has  the  greater  included 
angle. 

In  the  two  A's,  ABC  and 
A  CD,  let  AB  and  AC  of  the 
one  A  he  equal  to  AD  and  A  C 
of  the  other  A,  and  the  angle 
BAC  greater  than  the  angle 
DAC;  we  are  to  prove  that 
the  side  BCia  greater  than  the 
side  CD. 

Conceive  the  two  A's  joined  together  hy  their  shorter 
equal  sides,  and  draw  the  line  BD.  Now,  as  AB  =  AD, 
ABD  is  an  isosceles  A.  From  the  vertex  A,  draw  a  line 
bisecting  the  angle  BAD.  This  line  must  be  perpendic- 
ular to  the  base  BD,  (Th.  18,  Cor.  1).  Since  the  [_BAC 
is  greater  than  the  [_DAC,  this  line  must  meet  BC,  and 
will  not  meet  CD.  From  the  point  E,  where  the  per- 
pendicular meets  BC,  draw  BD. 

Now  BE  =  DE,  (Th.  18,  Scholium  1). 

Add  EC  to  each ;  then  BC=DE  +  EC. 

But  DE  +  EC  is  greater  than  DC. 

Therefore  BC>DC. 

Hence  the  theorem ;  if  two  triangles  have  two  sides  of 
one  equal  to  two  sides  of  the  other,  etc. 

Cor.  Any  point  out  of  the  perpendicular  drawn  from 
the  middle  point  of  a  line,  is  unequally  distant  from  the 
extremities  of  the  line. 


GEOMETRY. 


THEOREM    XXIII. 


A  perpendicular  is  the  shortest  line  that  can  be  drawn  from 
any  point  to  a  straight  line  ;  and  if  other  lines  be  drawn  from 
the  same  point  to  the  same  straight  line,  the  longer  line  will 
be  at  a  greater  distance  from  the  perpendicular;  and  lines  at 
equal  distances  from  the  perpendicular,  on  opposite  sides,  are 
equal. 

Let  A  be  any  point  without  the 
line  DE ;  let  AB  be  the  perpen- 
dicular; and  AC,  AD,  and  AE 
oblique  lines :  then,  if  BC  is  less 
than  BB,  and  BC=  BE,  we  are  to 
show, 

1st.  That  AB  is  less  than  AC. 
2d.  That  AC  is  less  than  AB.     3d.  That  A C=  AE. 

1st.  In  the  triangle  ABC,  as  AB  is  perpendicular  to 
BC,  the  angle  ABC  is  a  right  angle;  \__  C  +  [_BAC  = 
another  right  angle,  (Th.  11);  and  the  angle  BCA  is  less 
than  a  right  angle;  and,  as  the  greater  side  is  always 
opposite  the  greater  angle,  AB  is  less  than  AC;  and  AC 
may  be  any  line  not  identical  with  AB ;  therefore  a  per- 
pendicular is  the  shortest  line  that  can  be  drawn  from  A 
to  the  line  BE. 

2d.  As  the  two  angles,  ACB  and  ACT),  are  together 
equal  to  two  right  angles,  (Th.  1),  and  ACB  is  less  than 
a  right  angle,  ACB  must  be  greater  than  a  right  angle ; 
consequently,  the  [__  B  is  less  than  a  right  angle ;  and,  in 
the  A  ACB,  AB  is  greater  than  AC,  or  AC  is  less  than 

AD,  (Th.  19). 

3d.  In  the  A's  J.£Cand  ABE,  AB  is  common,  CB=* 
BE,  and  the  angles  at B are  right  angles ;  therefore,  AC  = 

AE,  (Th.  16). 

Hence  the  theorem ;  a  perpendicular  is  the  shortest  line, 
etc. 

Cor.  Conversely :  if  two  equal  oblique  lines  be  drawn 


BOOK  I.  39 

from  the  same  point  to  a  given  straight  line,  they  will 
meet  the  line  at  equal  distances  from  the  foot  of  the  per- 
pendicular drawn  from  that  point  to  the  given  line. 

THEOREM   XXIV. 

The  opposite  sides,  and  also  the  opposite  angles  of  any  par- 
allellogram,  are  equal. 

Let  ABOB  be  a  parallelogram. 
Then  we  are  to  show  that  AB  —  BO, 
AB  -  BO,  [_A  =  [_0,  and  \_ABQ 
=  \_ABO. 

Draw  a  diagonal,  as  BB ;  now,  be- 
cause AB  and  BO  are  parallel,  the  al- 
ternate angles  ABB  and  BBO  are  equal,  (Th.  6).     For 
the  same  reason,  as  AB  and  BO  are  parallel,  the  angles 
ABB  and  BBO  are  equal.     Now,  in  the  two  triangles 
ABB  and  BOB,  the  side  BB  is  common, 

the  [_ABB  =  [__BBO  (1) 

and  \__BBO  =  \_ABB  (2) 

Therefore,  the  angle  A  —  the  angle  O,  (Th.  11),  and  the 
two  A's  are  equal  in  all  respects,  (Th.  18) ;  that  is,  the 
sides  opposite  the  equal  angles  are  equal ;  or,  AB  =  BO, 
and  AB  =  BO.  By  adding  equations  ( 1 )  and  ( 2 ),  we  have 
the  angle  ABO=  the  angle  ABO,  (Ax.  2). 

Hence  the  theorem ;  the  opposite  sides,  and  the  opposite 
angles,  etc. 

Oor.  1.  As  the  sum  of  all  the  angles  of  the  quadrilateral 
is  equal  to  four  right  angles,  and  the  angle  A  is  always 
equal  to  the  opposite  angle  0;  therefore,  if  A  is  a  right 
angle,  0  is  also  a  right  angle,  and  the  figure  is  a  rect- 
angle. 

Oor.  2.  As  the  angle  ABO,  added  to  the  angle  A,  gives 
the  same  sum  as  the  angles  of  the  A  ABB;  therefore, 
the  two  adjacent  angles  of  a  parallelogram  are  together 
equal  to  two  right  angles.  This  corresponds  to  Th.  13, 
Cor.  2. 


40 


GEOMETRY. 


THEOREM   XXV. 

If  the  opposite  sides  of  a  quadrilateral  are  equal,  they  are 
also  parallel,  and  the  figure  is  a  parallelogram. 

Let  ABDO  be  any  quadrilateral; 
on  the  supposition  that  AD  =  BO,  and 
AB  =  DO,  we  are  to  prove  that  AD  is 
parallel  to  BO,  and  AB  parallel  to  DO. 

Draw  the  diagonal  BD;  we  now 
have  two  triangles,  ABD  and  BOD, 
which  have  the  side  BD  common,  AD  of  the  one  =  BO 
of  the  other,  and  AB  of  the  one  =  OD  of  the  other ; 
therefore  the  two  A's  are  equal,  (Th.  21),  and  the 
angles  opposite  the  equal  sides  are  equal ;  that  is,  the 
angle  ADB  =  the  angle  OBD ;  but  these  are  alternate 
angles;  and,  therefore,  AD  is  parallel  to  BO,  (Th.  7); 
and  because  the  angle  ABD  =  the  angle  BDO,  AB  is 
parallel  to  OD,  and  the  figure  is  a  parallelogram. 

Hence  the  theorem;  if  the  opposite  sides  of  a  quadri- 
lateral, etc. 

Oor.  This  theorem,  and  also  Th.  24,  proves  that  the 
two  A's  which  make  up  the  parallelogram  are  equal; 
and  thesame  would  be  true  if  we  drew  the  diagonal 
from  A  to  0;  therefore,  the  diagonal  of  any  parallelogram 
bisects  the  parallelogram. 


THEOREM   XXVI. 

The  lines  which  join  the  corresponding  extremities  of  two 
equal  and  parallel  strait  lines,  are  themselves  equal  and 
parallel ;  and  the  figure  thus  formed  is  a  parallelogram. 

On  the  supposition  that  AB  is 
equal  and  parallel  to  DO,  we  are  to 
prove  that  AD  is  equal  and  parallel 
to  BO;  and  that  the  figure  is  a  par- 
allelogram. 

Draw  the  diagonal  BD  ;  now,  since 


BOOK  I.  41 

AB  and  DC  are  parallel,  and  BB  joins  them,  the  alter- 
nate angles  ABB  and  BBC  are  equal ;  and  since  the 
side  AB  =  the  side  BO,  and  the  side  BB  is  common  to 
the  two  A's  ABB  and  OBB,  therefore  the  two  triangles 
are  equal,  (Th.  16) ;  that  is,  AB  =  BO,  the  angle  A  =  0, 
and  the  |_  ABB  =  the  \_BBO;  also  AB  is  parallel  to 
BO;  and  the  figure  is  a  parallelogram. 

Hence  the  theorem ;  the  lines  which  join  the  corresponding 
extremities,  etc. 

THEOREM   XXVII. 

Parallelograms  on  the  same  base,  and  between  the  same 
parallels,  are  equivalent,  or  equal  in  respect  to  area  or  sur- 
face. 

Let  ABEO  and  ABBF  be  two 
parallelograms  on  the  same  base 
AB,  and  between  the  same  paral- 
lels AB  and  OB  ;  we  are  to  prove 
that  these  two  parallelograms  are 
equal.  — 

!Nbw,  OB  and  FB  are  equal,  be- 
cause they  are  each  equal  to  AB,  (Th.  24) ;  and,  if  from 
the  whole  line  OB  we  take,  in  succession,  OB  and  FB, 
there  will  remain  EB  =  OF,  (Ax.  3) ;  but  BE  =  AO,  and 
AF=  BB,  (Th.  24);  hence  we  have  two  A's,  OAF  and 
EBB,  which  have  the  three  sides  of  the  one  equal  to  the 
three  sides  of  the  other,  each  to  each ;  therefore,  the  two 
A's  are  equal,  (Th.  21).  If,  from  the  whole  figure 
ABBO,  we  take  away  the  A  OAF,  the  parallelogram 
ABBF  will  remain ;  and  if  from  the  whole  figure  we  take 
away  the  other  A  EBB,  the  parallelogram  ABEO  will 
remain.  Therefore,  (Ax.  3),  the  parallelogram  ABBF  = 
the  parallelogram  ABEO. 

Hence  the  theorem ;  Parallelograms  on  the  same  base,  etc. 
4* 


42  GEOMETRY. 


THEOREM    XXVIII. 


Triangles  on  the  same  base  and  between  the  same  parallels 
are  equivalent. 

Let  the  two  a's  ABE  and  ABF     , 
have  the  same  base  AB,  and  be  be-     <T 
tween  the  same  parallels  AB  and       \ 
CD ;  then  we  are  to  prove  that  they 
are  equal  in  surface. 

From  B  draw  the  line  BD,  par- 
allel to  AF;  and  from  A  draw  the  line  AC,  parallel  to 
BE ;  and  produce  EF,  if  necessary,  to  C  and  D ;  now  the 
parallelogram  ABDF  ==  the  parallelogram  ABEC,  (Th. 
27).  But  the  A  ABE  is  one  half  the  parallelogram 
ABEC,  and  the  A  ABF  is  one  half  the  parallelogram 
ABDF;  and  halves  of  equals  are  equal,  (Ax.  7) ;  there- 
fore the  A  A  BE  =  the  A  ABF. 

Hence  the  theorem ;  triangles  on  the  same  base,  etc. 

THEOREM   XXIX. 

Parallelograms  on  equal  bases,  and  between  the  same  par- 
allels, are  equal  in  area. 

Let  ABCD  and  EFaH,  be  two      d 
parallelograms  on  equal  bases,  AB     j 
and  EF,   and  between  the    same     /  j> 
parallels,  AF  and  DCr ;  then  we  are    A 
to  prove  that  they  are  equal  in  area. 

AB  —  EF=EGr\  but  lines  which  join  equal  and 
parallel  lines,  are  themselves  equal  and  parallel,  (Th.  26) ; 
therefore,  if  AS  and  BGr  be  drawn,  the  figure  ABGffis 
•a  parallelogram  =  to  the  parallelogram  ABCD,  (Th.  27); 
and  if  we  turn  the  whole  figure  over,  the  two  parallelo- 
grams, GrHEF  and  GrEAB,  will  stand  on  the  same  base, 
CrH,  and  between  the  same  parallels ;  therefore,  GrHEF 
=  aHAB,  and  consequently  ABCD  =  EFGH,  (Ax.  1). 

Hence  the  theorem ;  Parallelograms  on  equal  bases,  etc. 


BOOK  X.  43 

Oor.  Triangles  on  equal  bases,  and  between  the  same 
parallels,  are  equal  in  area.  For,  draw BB  and  EG;  the 
A  ABB  is  one  half  of  the  parallelogram  AO,  and  the 
A  EFGr  is  one  half  of  the  equivalent  parallelogram  FE; 
therefore,  the  A  ABB  =  the  A  EFG,  (Ax.  7). 

THEOREM   XXX. 

If  a  triangle  and  a  parallelogram  are  upon  the  same  or  equal 
bases,  and  between  the  same  parallels,  the  triangle  is  equiva- 
lent to  one  half  the  parallelogram. 

Let  ABO  be  a  A,  and  ABBE  a 
parallelogram,  on  the  same  base  AB, 
and  between  the  same  parallels ;  then 
we  are  to  prove  that  the  A  ABO  is 
equivalent  to  one  half  of  the  parallel- 
ogram ABBE. 

Draw  the  diagonal  EB  to  the  parallelogram ;  now, 
because  the  two  A's  ABO  and  ABE  are  on  the  same 
base,  and  between  the  same  parallels,  they  are  equiva- 
lent, (Th.  28);  but  the  A  ABE  is  one  half  the  parallel- 
ogram ABBE,  (Th.  25,  Cor.) ;  therefore  the  A  ABO  is 
equivalent  to  one  half  of  the  same  parallelogram,  (Ax.  7). 

Hence  the  theorem ;  if  a  triangle  and  a  parallelogram, 
etc* 

THEOREM   XXXI. 

The  complementary  parallelograms  described  about  any 
point  in  the  diagonal  of  any  parallelogram,  are  equivalent  to 
each  other. 

Let  A  0  be  a  parallelogram,  and 
BB  its  diagonal ;  take  any  point, 
as  E,  in  the  diagonal,  and  through 
this  point  draw  lines  parallel  to  the 
sides  of  the  parallelogram,  thus 
forming  four  parallelograms. 

"We  are  now  to  prove  that  the  complementary  paral- 
lelograms, AE  and  EO,  are  equivalent. 


. 


44  GEOMETRY. 

By  (Th.  25,  Cor.)  we  learn  that  the  A  ABD  =  A  DBC. 
Also  by  the  same  Cor.,  A  a  =  A  b,  and  A  c=  A  d;  there- 
fore by  addition 

Now,  from  the  whole  A  ABD  take  A  a  +  A  c,  and 
from  the  whole  A  DBC  take  the  equal  sum,  A  b  -f-  A  d, 
and  the  remaining  parallelograms  AE  and  EC  are  equiv- 
alent, (Ax.  3). 

Hence  the  theorem ;  the  complementary  parallelograms, 
etc. 

THEOREM  XXXII. 

The  perimeter  of  a  rectangle  is  less  than  that  of  any  rhom- 
boid standing  on  the  same  base,  and  included  between  the  same 
parallels. 

Let  ABCD  be  a  rect- 
angle, and  ABEF&  rhom- 
boid having  the  same  base, 
and  their  opposite  sides 
in  the  same  line  parallel 
to  the  base. 

"We  are  now  to  prove  that  the  perimeter  ABODA  is  less 
than  ABEFA. 

Because  AD  is  a  perpendicular  from  A  to  the  line  DE, 
and  AF  an  oblique  line,  AD  is  less  than  AF,  (Th.  23). 
For  the  same  reason  BO  is  less  than  BE;  hence  AD  + 
BC< AF+  BE.  Adding  the  sum,  AB  +  DO,  to  the  first 
member  of  this  inequality,  and  its  equal  AB  -f  FE  to 
the  second  member,  we  have  AB  +  BC  +  CD  +  DA,  or 
the  perimeter  of  the  rectangle,  less  than  AB  -f  BE  + 
EF  +  FA,  or  the  perimeter  of  the  rhomboid.  Hence 
the  theorem ;  the  perimeter  of  a  rectangle,  etc. 

Scholium. — In  Theorem  30  it  is  shown  that  the  triangles  ABC,  ABE, 
and  DBE,  are  equal  in  area,  and  that  each  is  equal  to  one  half  the 
parallelogram  ABBE.  This  parallelogram  also  has  the  same  area  as 
the  rectangle  having  an  equal  base  and  altitude. 


BOOK  I. 


45 


Thus  far,  areas  have  been  considered  only  relatively 
and  in  the  abstract.  We  will  now  explain  how  we  may 
pass  to  the  absolute  measures,  or,  more  properly,  to  the 
numerical  expressions  for  areas. 

THEOREM  XXXIII. 

The  area  of  any  plane  triangle  is  measured  by  the  product 
of  its  base  by  one  half  its  altitude;  or  one  half  its  base  by 
its  altitude,  or  one  half  the  product  of  its  base  by  its  altitude. 

Let  ABO  represent  any  triangle,  AB 
its  base,  and  AD,  at  right  angles  to  AB, 
its  altitude ;  now  we  are  to  show  that  the 
area  of  ABC  is  equal  to  the  product  of 
AB  by  one  half  of  AD ;  or  one  half  of    *  * 

AB  by  AD ;  or  one  half  of  the  product  of  AB  by  AD. 

On  AB  construct  the  rectangle  ABED',  and  the  area 
of  this  rectangle  is  measured  by  AB  into  AD  (Def. 
54) ;  but  the  area  of  the  A  ABO  is  equivalent  to  one 
half  this  rectangle,  (Th.  30).  Therefore,  the  area  of  the 
A  is  measured  by  J  AB  x  AD,  or  one  half  the  product 
of  its  base  by  its  altitude.  Hence  the  theorem ;  the  area 
of  any  plane  triangle,  etc. 

THEOREM  XXXIV. 

The  area  of  a  trapezoid  is  measured  by  one  half  the  sum 
of  its  parallel  sides  multiplied  by  the  perpendicular  distance 
between  them. 

JjetABDO  represent  any  trape- 
zoid; draw  the  diagonal  BO,  divid- 
ing it  into  two  triangles,  ABO  and 
BOD:  OD  is  the  base  of  one  tri- 
angle, and  AB  may  be  considered 
as  the  base  of  the  other ;  and  EF  is  the  common  altitude 
of  the  two  triangles. 

Now,  by  Th.  33,  the  area  of  the  triangle  BOD  =  \  OD 
x  EF;  and  the  area  of  the  A  ABC=  \AB  x  EF;  but 


46 


GEOMETRY. 


L    K      I 


by  addition,  the  area  of  the  two  A's,  or  of  the  trape- 
zoid, is  equal  to  J  ( AB+  CD)  x  EF.  Hence  the  theorem ; 
the  area  of  a  trapezoid,  etc. 

THEOREM  XXXV. 

If  one  of  two  lines  is  divided  into  any  number  of  parts,  the 
rectangle  contained  by  the  two  lines  is  equal  to  the  sum  of  the 
several  rectangles  contained  by  the  undivided  line  and  the  seve- 
ral parts  of  the  divided  line. 

Let  AB  and  AD  be  two  lines, 
and  suppose  AB  divided  into  any 
number  of  parts  at  the  points  E, 
F,  Gr,  etc. ;  then  the  whole  rect- 
angle contained  by  the  two  lines 
is  AH,  which  is  measured  by  AB      A  *     *     u      u 

into  AB.  But  the  rectangle  AL  is  measured  byf  AE 
into  AD ;  the  rectangle  EK  is  measured  by  EF  into  EL, 
which  is  equal  to  EF  into  AD ;  and  so  of  all  the  other 
partial  rectangles ;  and  the  truth  of  the  proposition  is  as 
obvious  as  that  a  whole  is  equal  to  the  sum  of  all  its 
parts.  Hence  the  theorem ;  if  one  of  two  lines  is  divided, 
etc. 


THEOREM    XXXVI. 

If  a  straight  line  is  divided  into  any  two  parts,  the  square 
described  on  the  whole  line  is  equivalent  to  the  sum  of  the 
squares  described  on  the  two  parts  plus  twice  the  rectangle  con 
tained  by  the  parts. 

Let  AB  be  any  Hue  divided  into 
any  two  parts  at  the  point  C;  now  we 
are  to  prove  that  the  square  on  AB 
is  equivalent  to  the  sum  of  the 
squares  on  A  C  and  CB  plus  twice  the 
rectangle  contained  by  AC  and  CB. 

On  AB  describe  the  square  AD. 
Through  the  point  C  draw  CM,  par- 


BOOK    I.  47 

allel  to  BB ;  take  BE  m  BO,  and  through  E  draw  EKN, 
parallel  to  AB.  We  now  have  OE,  the  square  on  CB,  by 
direct  construction. 

As  AB  mm  BB,  and  OB  =  BE,  by  subtraction,  AB  — 
CB  =  BB  —  BE;  or  AC =  EB.  But  NK=  AC,  being 
opposite  sides  of  a  parallelogram ;  and  for  the  same  rea- 
son, KM  =  EB.  Therefore,  (Ax.  1),  NK  =  KM,  and  the 
figure  NM  is  a  square  on  NK,  equal  to  a  square  on  A  0. 
But  the  whole  square  on  AB  is  composed  of  the  two 
squares  CE,  NM,  and  the  two  complements  or  rectangles 
u4.iT and  KB-,  and  each  of  these  latter  is  ACm  length, 
and  BO  in  width ;  and  each  has  for  its  measure  AO  into 
OB ;  therefore  the  whole  square  on  AB  is  equivalent  to 
AC2  +  BC2  +  2AC  x  OB. 

Hence  the  theorem ;  if  a  straight  line  is  divided  into  any 
two  parts,  etc. 

This  theorem  may  be  proved  algebraically,  thus : 

Let  w  represent  any  whole  right  line  divided  into  any 
two  parts  a  and  b ;  then  we  shall  have  the  equation 

ic  =  a  -f  b 
By  squaring,  w2  =  a2  -f  b2  -f  2ab. 

Oor.  If  a  =  b,  then  w2  —  4a2 ;  that  is,  the  square  de- 
scribed on  any  line  is  four  times  the  square  described  on 
one  half  of  it. 

THEOREM    XXXVII. 

The  square  described  on  the  difference  of  two  lines  is  equiv- 
alent to  the  sum  of  the  squares  described  on  the  two  lines  di- 
minished by  twice  the  rectangle  contained  by  the  lines. 

Let  AB  represent  the  greater  of  two  lines,  OB  the 
less  line,  and  A 0  their  difference. 

We  are  now  to  prove  that  the  square  described  ow  AC 
is  equivalent  to  the  sum  of  the  squares  on  AB  and  BC 
diminished  by  twice  the  rectangle  contained  by  AB 
and  BO. 

Conceive  the  square  AF  to  be  described  on  AB,  and 


48  GEOMETRY. 

the  square  BL  on  CB ;  on  A  C  describe 
the  square  ACGM,  and  produce  MG 
to  K. 

As  GC=AC,  and  CL  =  CB,  by 
addition,  (GC  +  CL),  or  GL,  is  equal 
to  AO+  CB,  or  4.B.  Therefore,  the 
rectangle  GE  is  J.1?  in  length,  and 
CB  in  width,  and  is  measured  by  AB 

XBC. 
Also  AE=  AB,  and  AM=  AC;  hy  subtraction,  MH 

=  CB;  and  as  MK=  AB,  the  rectangle  IZjKT  is  AB  in 
length,  and  Ci?  in  width,  and  is  measured  by  AB  X  BC; 
and  the  two  rectangles  GE  and  HK  are  together  equiva- 
lent to  2AB  x  BC. 

]$o w,  the  squares  on  .Ai?  and  BC  make  the  whole  figure 
AHFELC;   and  from  this  whole  figure,  or  these  two 
squares,  take  away  the  two  rectangles  iTiT  and  GE,  and 
the  square  on  A  C  only  will  remain ;  that  is, 
AC2=AB2  +  BC  —  2AB  x  BC 

Hence  the  theorem ;  the  square  described  on  the  differ- 
ence of  two  lines,  etc. 

This  theorem  may  be  proved  algebraically,  thus : 

Let  a  represent  the  greater  of  two  lines,  b  the  less,  and 
d  their  difference ;  then  we  must  have  this  equation : 
d  =  a  — b 
By  squaring,    d2  =  a2  -f  b2  —  2ab. 

a  a2 

Cor.  If  d=  b,  then  d  =  «",  and  d2  =  -r ;    that  is,   the 

square  described  on  one  half  of  any  line  is  equivalent 
to  one  fourth  of  the  square  described  on  the  whole  line. 

THEOREM   XXXVIII. 

The  difference  of  the  squares  described  on  any  two  lines  is 
equivalent  to  the  rectangle  contained  by  the  sum  and  difference 
of  the  lines. 

Let  AB  be  the  greater  of  two  lines,  and  AC  the  less, 
and  on  these  lines  describe  the  squares  AD,  AM;  then,  the 


BOOK   I.  49 

difference  of  the  squares  on  AB  and  AC  is  the  two  rect- 
angles EF  and  FC.     We  are  now  to 
show  that  the  measure  of  these  rect- 
angles may  be  expressed  by  (A B  +  AC) 
x(AB  —  AC). 

The  length  of  the  rectangle  EF  is  ED, 
or  its  equal  AB;  and  the  length  of  the 
rectangle  FC  is  MC,  or  its  equal  AC; 
therefore,  the  length  of  the  two  together  (if  we  con- 
ceive them  put  between  the  same  parallel  lines)  will  be 
AB  +  AC;  and  the  common  width  is  CB,  which  is  equal 
to  AB—AC;  therefore,  AB2— -AC2=  (AB+AC)  x  (AB 
--AC). 

Hence  the  theorem;  the  difference  of  the  squares  de- 
scribed on  any  two  lines,  etc. 

This  theorem  may  be  proved  algebraically:  thus, 

Let  a  represent  one  line,  and  b  another ; 

Then  a  -f  b  is  their  sum,  and  a  —  b  their  difference ; 

and  (a  +  b)  X  (a  —  b)  =  a2  —  b\ 

THEOREM   XXXIX. 

The  square  described  on  the  hypotenuse  of  any  right-angled 
triangle  is  equivalent  to  the  sum  of  the  squares  described  on 
the  other  two  sides. 

Let  ABC  represent  any  right-angled  triangle,  the  right 
angle  at  B;  we  are  to  prove  that  the  square  on  A  C  is 
equivalent  to  the  sum  of  two  squares;  one  on  AB,  the 
other  on  BC. 

On  the  three  sides  of  the  triangle  describe  the  three 
squares,  AB,  AL  and  BM.  Through  the  point  B,  draw 
BNE  perpendicular  to  AC,  and  produce  it  to  meet  the 
line  QI  in  K;  also  produce  AF  to  meet  Gfl  in  H,  and 
ML  to  meet  the  point  in  K. 

Remark.  —  That  the  lines,  GI  and  ML,  produced,  meet  at  the  point 
K,  may  be  readily  shown.     As  the  proof  of  this  fact  is  not  necessary  for 
the  demonstration,  it  is  left  afl  an  exercise  for  the  learner. 
5  D 


50 


GEOMETRY. 


The  angle  BAG  is  a  right  angle,  and  the  angle  NAE 
is  also  a  right  angle ;  if 
from  these  equals  we 
subtract  the  common 
angle  BAR,  the  re- 
maining angle,  BAG, 
must  be  equal  to  the  re- . 
maining  angle  GAH. 
The  angle  G  is  a  right 
angle,  equal  to  the 
angle  ABO;  and  AB 
=  AG ;  therefore,  the 
two  A's  ABC  and 
AGH  are  equal,  and 
AH=AC.  ButA(7  = 
AF;  therefore,  AH= 
AF.      Now,   the    two 

parallelograms,  AF  and  ARKB  are  equivalent,  because 
they  are  upon  equal  bases,  and  between  the  same  paral- 
lels, FH  and  FK,  (Th.  27). 

But  the  square  A  I,  and  the  parallelogram  AHKB,  are 
equivalent,  because  they,  are  on  the  same  base,  AB,  and 
between  the  same  parallels,  AB  and  GK;  therefore,  the 
square  Al,  and  the  parallelogram  AF,  being  each  equiv- 
alent to  the  same  parallelogram  AHKB,  are  equivalent 
to  each  other,  (Ax.  1).  ,In  the  same  manner  we  may 
prove  that  the  square  BD  is  equivalent  to  the  rectangle 
ND ;  therefore,  by  addition,  the  two  squares,  A I  and 
BM,  are  equivalent  to  the  two  parallelograms,  AF  and 
ND,  or  to  the  square  AD. 

Hence  the  theorem ;  the  square  described  on  the  hypote- 
nuse of  a  right-angled  triangle,  etc. 

Cor.  If  two  right-angled  triangles  have  the  hypotenuse,  and 
a  side  of  the  one  equal  to  the  hypotenuse  and  a  side  of  the 
other,  each  to  each,  the  two  triangles  are  equal. 


BOOK  I.  51 

Let  ABC  and  AGHbe  the  two  A's,  in  which  we  sup- 
pose ACfam  AH,  and  BC  =  GH;  then  will  AG  =  AB. 

For,  we  have  AC2  =  AB2  +  BO2, 

or,  by  transposing,  AC2  —  BC*  =  J.J52, 

and  AH2  =  ~AG2  +  GH2, 

or,  by  transposing,  AH2 —  GH2  =  AG2. 

But  by  the  hypothesis  AC2  —~BC2  =  AH2  —  (T^2 ; 

hence,    AB2  =  J.  G\  or,  Jl J5  =  A  G. 

Scholium. — The  two  sides,  AB&nd  BC,  may  vary,  while  AC  remains 
constant.  AB  may  be  equal  to  BC;  then  the  point  iVwill  be  in  the 
middle  of  A  C.  When  AB  is  very  near  the  length  of  A  C,  and  BC  very 
small,  then  the  point  N  falls  very  near  to  C.  Now  as  AE  and  AD  are 
right-angled  parallelograms,  their  areas  are  measured  by  the  product 
of  their  bases  by  their  altitudes ;  and  it  is  evident  that,  as  they  have  the 
same  altitude,  these  areas  will  vary  directly  as  their  bases  AN  and 
NC;  hence  the  squares  on  AB  and  BC,  which  are  equivalent  to  those 
rectangles,  vary  as  the  lines  AN  and  NC. 

The  following  outline  of  the  demonstration  of  this  pro- 
position is  presented  as  a  useful  disciplinary  exercise  for 
the  student. 

We  employ  the  same  figure,  in  which  no  change  is 
made  except  to  draw  through  0  the  line  CP,  parallel  to  BK. 

The  first  step  is  to  prove  the  equality  of  the  triangles 
AGE  and  ABO,  whence  AH =  AC.  But  AC  =  AF; 
therefore  AH=  AF. 

The  parallelograms  AFEN  and  AHKB  are  equiva- 
lent. Also,  the  parallelogram  AHKB  =  the  square  ABIG, 
(Th.  27),  and  the  parallelogram  KBCP=NEDC=  square 
BCML.  .  Now,  by  adding  the  equals 
AFEN=  ABIG 
NEBC =  BCML 

we  obtain  AFD  C  =  ABIG  -f  B  CML. 

That  is,  the  square  on  A  C  is  equivalent  to  the  sum  of 
the  squares  on  AB  and  BC. 

The  great  practical  importance  of  this  theorem,  in  the 
extent  and  variety  of  its  applications,  and  the  frequency 
of  its  use  in  establishing  subsequent  propositions,  ren- 
der it  necessary  that  the  student  should  master  it  com- 
pletely.    To  secure  this  end,  we  present  a 


52 


GEOMETRY. 


Second  Demonstration, 

Let  ABO  be  a  triangle 
right-angled  at  B.  On  the 
hypotenuse  A  0,  describe  the 
square  A  CUB.  From  B  and 
E  let  fall  the  perpendiculars 
Bb  and  Ed,  on  AB  and  A B 
produced.  Draw  Bn  and  Oa, 
making  right  angles  with 
Ed. 

"We  give  an  outline  only 
of  the  demonstration,  requiring  the  pupil  to  make  it 
complete. 

First  Part. — Prove  the  four  triangles  ABO,  AbB,  BnE, 
and  EaO,  equal  to  each  other. 

The  proof  is  as  follows:  The  A's  ABO  and  BnE  are 
equal,  because  the  angles  of  the  one  are  equal  to  the 
angles  of  the  other,  each  to  each,  and  the  hypotenuse 
AO  of  the  one,  is  equal  to  the  hypotenuse  BE  of  the 
other.  In  like  manner,  it  may  be  shown  that  the  a's 
AbB  and  EaO  are  equal. 

Now,  the  sum  of  the  three  angles  about  A,  is  equal  to 
the  sum  of  the  three  angles  of  the  A  ABO)  and  if,  from 
the  first  sum,  we  take  [_BAO  -f  L_  CAB,  and  from  the 
second  we  take  L^  +  L CAB  =  [_BAO+  [__  OAB,  the 
remaining  angles  are  equal ;  that  is,  [_  Bb  A  is  equal  to 
[_AOB ;  hence  the  A's  ABO  and  BbA  have  their  angles 
equal,  each  to  each;  and  since  AO  —  BA,  the  A's  are 
themselves  equal,  and  the  four  triangles  ABO,  AbB, 
BnE,  and  EaO,  are  equal  to  each  other. 

Second.  —  Prove  that  the  square  bBnd  is  equal  to  a 
square  on  AB.     The  square  BdaCis  obviously  on  BO. 

Third. — The  area  of  the  whole  figure  is  equal  to  the 
square  on  AO,  and  the  area  of  two  of  the  four  equal 
right-angled  triangles. 

Also,  the  area  of  the  whole  figure  is  equal  to  two  other 


BOOK   1.  53 

squares,  bDnd  and  daOB,  and  two  of  the  four  equal  tri- 
angles, DnE  and  EaO. 

Omitting  or  subtracting  the  areas  of  two  of  the  four 
right-angled  triangles,  in  each  of  the  two  expressions  for 
the  area  of  the  whole  figure,  there  will  remain  the  square 
on  A  0,  equal  to  the  sum  of  the  two  squares,  Dndb  and 
da  OB. 


"2       .       TT7v2  -T-F*1 


That  is,  AB'  +  BO    =  AG 

Hence  the  theorem;  the  square  described  on  the  hypote- 
nuse of  a  right-angled  triangle,  etc. 

Scholium. — Hence,  to  find  the  hypotenuse  of  a  right-angled  triangle, 
extract  the  square  root  of  the  sum  of  the  squares  of  the  two  sides  about 
the  right  angle. 

THEOREM    XL. 

In  any  obtuse-angled  triangle,  the  square  on  the  side  oppo- 
site the  obtuse  angle  is  greater  than  the  sum  of  the  squares 
on  the  other  two  sides,  by  twice  the  rectangle  contained  by 
.either  side  about  the  obtuse  angle,  and  the  part  of  this  side 
produced  to  meet  the  perpendicular  drawn  to  it  from  the 
vertex  of  the  opposite  angle. 

Let  ABO  be  any  triangle  in  which 
the  angle  at  B  is  obtuse.  Produce 
either  side  about  the  obtuse  angle, 
as  OB,  and  from  A  draw  AD  perpen- 
dicular to  OB,  meeting  it  produced 
atD. 

It  is  obvious  that  OD  =  OB  +  BD. 
By  squaring,  ~OD2  =  OB2  +  20B  x  BD  +  BD2,  (Th.  36). 
Adding  AD2  to  each  member  of  this  equation,  we  have 
AD2  +OD2  =  OB2  +  BD2  -f  AD2  +  20B  x  BD. 

But,  (Th.  39),  the  first  member  of  the  last  equation  is 
equal  to  AO2,  and 


BD2  +  AD2  =  AB\ 

5* 


54  GEOMETRY. 

Therefore,  this  equation  becomes 

~AC2  -  CB2  +  AB*  +  2CB  x  BD. 

That  is,  the  square  on  A  0  is  equivalent  to  the  sum  of 
the  squares  on  CB  and  AB,  increased  by  twice  the  rect- 
angle contained  by  CB  and  BD. 

Hence  the  theorem;  in  any  obtuse -angled  triangle,  the 
square  on  the  side  opposite  the  obtuse  angle,  etc. 

Scholium. — Conceive  AB  to  turn  about  the  point  A,  its  intersection 
with  CD  gradually  approaching  D.  The  last  equation  above  will  be 
true,  however  near  this  intersection  is  to  D,  and  when  it  falls  upon  D 
the  triangle  becomes  right-angled. 

In  this  case  the  line  BD  reduces  to  zero,  and  the  equation  becomes 
AC7=  CB*  +  AB*,  in  which  CB  and  AB  are  now  the  base  and  per- 
pendicular of  a  right-angled  triangle.  This  agrees  with  Theorem  39, 
as  it  should,  since  we  used  the  property  of  the  right-angled  triangle 
established  in  Theorem  39  to  demonstrate  this  proposition ;  and  in  the 
equation  which  expresses  a  property  of  the  obtuse-angled  triangle,  we 
have  introduced  a  supposition  which  changes  it  into  one  which  is 
right-angled. 

THEOREM   XLI. 

In  any  triangle,  the  square  on  a  side  opposite  an  acute  angle 
is  less  than  the  sum  of  the  squares  on  the  other  two  sides,  by 
twice  the  rectangle  contained  by  either  of  these  sides,  and  the 
distance  from  the  vertex  of  the  acute  angle  to  the  foot  of  the 
perpendicular  let  fall  on  this  side,  or  side  produced,  from  the 
vertex  of  its  opposite  angle. 

Let  ABC,  either 
figure,  represent 
any  triangle ;  C  an 
acute  angle,  CB 
the  base,  and  AB 
the  perpendicular, 
which  falls  either 
without  or  on  the  base.    !Now  we  are  to  prove  that 

AB>=  CB2  +  AC2—  2CB  x  CD. 


HOOK     I 


55 


From  the  first  figure  we  get  BB=-CB—OB        ( 1 ) 
and  from  the  second  BB -  OB  —OB        ( 2 ) 

Either  ono  of  these  equations  will  give,  (Th.  87), 

BB2 -  OB2  +  OB2—20B  x  OB. 

Adding  AD*  to  each  member  and  reducing,  we  obtain, 
(Th.  89),  AB2  ~Jd2+OB2  —  20Bx  OB,  which  proves 
the  proposition.    Hence  the  theorem. 


THEOREM   XLII. 

If  in  any  triangle  a  line  be  drawn  from  any  angle  to  the 
middle  of  the  opposite  Me,  twice  the  square  of  this  line, 
together  with  twice  the  square  of  one  half  the  side  bisected,  will 
be  equivalent  to  the  sum  of  the  squares  of  the  other  two  sides. 

Let  AB 0  be  a  triangle,  and 
M  the  middle  point  of  its 
base. 

Then  we  are  to  prove  that 
2AM2  +  20M a  -  A02+ AJB*. 

Draw  AB  perpendicular  to 
the  base,  and  make  AB  «  p} 
AO=b,  AB=*c,  OB**2a} 
AM  =  m,  and  MB  =■  x;  then  OM  —  a,  OB  =»  a  +  z,  BB 

e»  a  —  X. 

Now  by,  (Th.  39),  wo  have  the  two  following  equations : 

p'  +  (a  —  zy  =  c*  (1) 

f  +  {a  +  xy  »  y  (2) 

By  addition,  2tf  +  2x%  +  2a%  m  6"  -f  c\   But  f  -f  x7  —  m\ 

Therefore,  2m9  +  2a*  -  b2  +  c\ 

This  equation  is  tho  algebraic  enunciation  of  the 
theorem. 


56  GEOMETRY. 


THEOREM    XLIII. 


The  two  diagonals  of  any  parallelogram  bisect  each  other  ; 
and  the  sum  of  their  squares  is  equivalent  to  the  sum  of  the 
squares  of  the  four  sides  of  the  parallelogram. 

I^etABCD  be  any  parallelogram, 
and  A  0  and  BD  its  diagonals. 

We  are  now  to  prove, 

1st.  That  AE  mm  EC,  and  DE  = 

m 

2d.  That  AC2  +  'BD2  =  AB*  +  ~BC2  +  CD2  +~AD\ 

1.  The  two  triangles  ABU  and  CDE  are  equal,  be- 
cause AB  =  CD,  the  angle  ABE  =  the  alternate  angle 
ODE,  and  the  vertical  angles  at  E  are  equal ;  therefore, 
AE,  the  side  opposite  the  angle  ABE,  is  equal  to  CE, 
the  side  opposite  the  equal  angle  CDE;  also  EB,  the 
remaining  side  of  the  one  A,  is  equal  to  EB,  the  remain- 
ing side  of  the  other  triangle. 

2.  As  AOB  is  a  triangle  whose  base,  AC,  is  bisected 
in  E,  we  have,  by  (Th.  42), . 


2AK  +  2ED'  =  AD*  +  DO2      ( 1 ) 

And  as  A  OB  is  a  triangle  whose  base,  AG,  is  bisected 
in  E,  we  have 


2AE'  +  2EB  =  AB'  +  BQ*     ( 2 ) 
By  adding  equations  (1)  and  (2),  and  observing  that 
EB2  =  ED2,  we  have 
±AE2  +  4ED2  =  ~AD2  +~DC2  +  AB2  +~BC2 

But,  four  times  the  square  of  the  half  of  a  line  is  equiv- 
alent to  the  square  of  the  whole  line,  (Th.  36,  Corollary) ; 
therefore  4 AE2  =  AC2,  and  4ED2  =  DW;  and  by  sub- 
stituting these  values,  we  have 


AC  +  BD'  -  AB'  +  £<T  +  DC  +  .42r, 
which   equation    conforms  to  the    enunciation  of  the 
theorem. 


BOOK   I 


57 


C                 H 

L 

M 


THEOREM   XLIV. 

If  a  line  be  bisected  and  produced,  the  rectangle  contained 
by  the  whole  line  and  the  part  produced,  together  with  the 
square  of  one  half  the  bisected  line,  will  be  equivalent  to  the 
square  on  a  line  made  up  of  the  line  produced  and  one  half  the 
bisected  line,    ^ 

Let  AB  be  any  line,  bi- 
sected in  0  and  produced 
to  2).  On  CD  describe 
the  square  OF,  and  on 
BD  describe  the  square 
BE. 

The  sides  of  the  square 
BE  being  produced,  the 
square  GrL  will  be  form- 
ed.   Also,  complete  the  construction  of  the  rectangle 
ADEK. 

Then  we  are  to  prove  that  the  rectangle,  AE,  and  the 
square,  GrL,  are  together  equivalent  to  the  square, 
CDFa. 

The  two  complementary  rectangles,  CL  and  LF,  are 
equal,  .(Th.  31).  But  CL=AH,  the  line  AB  being  bisected 
at  0;  therefore  AL  is  equal  to  the  sum  of  the  two  com- 
plementary rectangles  of  the  square  CF.  To  AL  add 
the  square  BE,  and  the  whole  rectangle,  AE,  will  be 
equal  to  the  two  rectangles  OE  and  EM.  To  each  of 
these  equals  add  HM,  or  the  square  on  HL  or  its  equal 
OB,  and  we  have  rectangle  AE  -f  square  HM  =  CD2 ; 
but  rectangle  AE  =  AD  x  BD,  and  square  HM  =  CB2* 
Hence  the  theorem,  etc. 

Scholium.  —  If  we  represent  AB  by  2a,  and  BD  by  x,  then  AD  —■ 
2a +  x,  and  AD  X  BD  =  203^^.  But  CW  =  a2;  adding  this 
equation  to  the  preceding,  member  to  member,  we  get  AD  X  BD  -f- 
CB2  =  a?  +  2ax  +  x*  =  a  +  x.  But  CD  — a  +  x;  hence  this  equa- 
tion is  equivalent  to  the  equation  AD  X  DB  -f-  CB2  =  CD ,  which  is 
the  algebraic  proof  of  the  theorem. 


•l  «u.-L  •+-  * 


58  GEOMETRY. 

THEOREM  XLV. 

If  a  straight  line  be  divided  into  two  equal  parts,  and  also 
into  two  unequal  parts,  the  rectangle  contained  by  the  two  un- 
equal parts  together  with  the  square  of  the  line  between  the 
points  of  division,  will  be  equivalent  to  the  square  on  one  half 
the  line. 

Let  AB  be  a  line  bisected  in  C,  and  divided  into  two 
unequal  parts  in  D. 

We  are  to  prove 

that  AB  x  BB  +         - g c 

C52  =  AC\  orm\  :  A 

We  see  by  inspection  that  AB  —  AC  +  CB,  and  BB 
m  AC —  CB;  therefore  by  multiplication  we  have 
ABxBB  =  AC2  —~CB2,  (Th.  38). 

By  adding  CB2  to  each  of  these  equals,  we  obtain 
AB  x  BB +~CB2  =~AC2 

Kence  the  theorem. 


BOOK   II.  59 


BOOK   II. 


PROPORTION. 
DEFINITIONS  AND  EXPLANATIONS. 

The  word  Proportion,  in  its  common  meaning,  de- 
notes that  general  relation  or  symmetry  existing  between 
the  different  parts  of  an  object  which  renders  it  agree- 
able to  our  taste,  and  conformable  to  our  ideas  of  beauty 
or  utility ;  but  in  a  mathematical  sense, 

1.  Proportion  is  the  numerical  relation  which  one  quan- 
tity bears  to  another  of  the  same  kind. 

As  the  magnitudes  compared  must  be  of  the  same  kind, 
proportion  in  geometry  can  be  only  that  of  a  line  to  a 
line,  a  surface  to  a  surface,  an  angle  to  an  angle,  or  a  volume 
to  a  volume, 

2.  Ratio  is  a  term  by  which  the  number  which  meas- 
ures the  proportion  between  two  magnitudes  is  desig- 
nated, and  is  the  quotient  obtained  by  dividing  the  one 

by  the  other.     Thus,  the  ratio  of  A  to  B  is  - ,  or  A  :  B, 

in  which  A  is  called  the  antecedent,  and  B  the  consequent. 
If,  therefore,  the  magnitude  A  be  assumed  as  the  unit  or 
standard,  this  quotient  is  the  numerical  value  of  B  ex- 
pressed in  terms  of  this  unit. 

It  is  to  be  remarked  that  this  principle  lies  at  the  found- 
ation of  the  method  of  representing  quantities  by  num- 
bers. For  example,  when  we  say  that  a  body  weighs 
twenty-five  pounds,  it  is  implied  that  the  weight  of  this 
body  has  been  compared,  directly  or  indirectly,  with  that 
of  the  standard,  one  pound.    And  so  of  geometrical 


60  GEOMETRY. 

magnitudes ;  when  a  line,  a  surface,  or  a  volume  is  said 
to  be  fifteen  linear,  superficial,  or  cubical  feet,  it  is  un- 
derstood that  it  has  been  referred  to  its  particular  unit, 
and  found  to  contain  it  fifteen  times ;  that  is,  fifteen  is 
the  ratio  of  the  unit  to  the  magnitude. 

"When  two  magnitudes  are  referred  to  the  same  unit, 
the  ratio  of  the  numbers  expressing  them  will  be  the 
ratio  of  the  magnitudes  themselves. 

Thus,  if  J.  and  B  have  a  common  unit,  a,  which  is 
contained  in  A,  m  times,  and  in  B,  n  times,  then  A  =  ma 

*  i>  A  B       na       n 

and  B  =  na,  and  —  =  —  =  -. 
A      ma      m 

To  illustrate,  let  the 

line  A  contain  the  line  A 

a  six  times,  and  let  the  t        t 

line  B  contain  the  same  a 

line  a  five  times :    then 

I j i j i I 

A=6a  and  B—5a,  which  B 

.      B      5a-     5 
glYeA=6a  =  6- 

3.  A  Proportion  is  a  formal  statement  of  the  equality 
of  two  ratios. 

Thus,  if  we  have  the  four  magnitudes  A,  B,  0  and  i>, 

such  that  —  =  — ,  this  relation  is  expressed  by  the  pro- 
portion A  :  B  : :  0 :  D,  or  A  :  B  =  0 :  2>,  the  first  of 
which  is  read,  A  is  to  B  as  0  is  to  B ;  and  the  second, 
the  ratio  of  A  to  B  is  equal  to  that  of  0  to  D. 

4.  The  Terms  of  a  proportion  are  the  magnitudes,  or 
more  properly  the  representatives  of  the  magnitudes 
compared. 

5.  The  Extremes  of  a  proportion  are  its  first  and  fourth 
terms. 

6.  The  Means  of  a  proportion  are  its  second  and  third 
terms. 

7.  A  Couplet  consists  of  the  two  terms  of  a  ratio*    The 


BOOK   II.  61 

first  and  second  terms  of  a  proportion  are  called  the 
first  couplet,  and  the  third  and  fourth  terms  are  called 
the  second  couplet. 

8.  The  Antecedents  of  a  proportion  are  its  first  and 
third  terms. 

9.  The  Consequents  of  a  proportion  are  its  second  and 
fourth  terms. 

In  expressing  the  equality  of  ratios  in  the  form  of  a 
proportion,  we  may  make  the  denominators  the  ante- 
cedents, and  the  numerators  the  consequents,  or  the 
reverse,  without  affecting  the  relation  between  the  magni- 
tudes. It  is,  however,  a  matter  of  some  little  importance 
to  the  beginner  to  adopt  a  uniform  rule  for  writing  the 
terms  of  the  ratios  in  the  proportion ;  and  we  shall  always, 
unless  otherwise  stated,  make  the  denominators  of  the 
ratios  the  antecedents,  and  the  numerators  the  conse- 
quents.* 

10.  Equimultiples  of  magnitudes  are  the  products  arising 
from  multiplying  the  magnitudes  by  the  same  number. 
Thus,  the  products,  Am  and  Bm,  are  equimultiples  of 
A  and  B. 

U.  A  Mean  Proportional  between  two  magnitudes  is  a 
magnitude  which  will  form  with  the  two  a  proportion, 
when  it  is  made  a  consequent  to  the  first  ratio,  and  an 
antecedent  to  the  second.  Thus,  if  we  have  three  mag- 
nitudes A,  B,  and  (7,  such  that  A  :  B  : :  B  :  (7,  B  is  a 
mean  proportional  between  A  and  O. 

12.  Two  magnitudes  are  reciprocally,  or  inversely  pro- 
portional when,  in  undergoing  changes  in  value,  one  is 
multiplied  and  the  other  is  divided  by  the  same  number. 
Thus,  if  A  and  B  be  two  magnitudes,  so  related  that  when 

B 

A  becomes  mA,  B  becomes  — ,  A  and  B  are  said  to  be 

m 

inversely  proportional. 

*  For  discussion  of  the  two  methods  of  expressing  Katio,  see  Uni- 
versity Algebra. 

6 


62  GEOMETRY. 

13.  A  Proportion  is  taken  inversely  when  the  ante- 
cedents are  made  the  consequents  and  the  consequents 
the  antecedents.     Thus 

14.  A  Proportion  is  taken  alternately,  or  by  alternation, 
when  the  antecedents  are  made  one  couplet  and  the  con- 
sequents the  other. 

15.  Mutually  Equiangular  Polygons  have  the  same  num- 
ber of  angles,  those  of  the  one  equal  to  those  of  the 
others,  each  to  each,  and  the  angles  like  placed. 

16.  Similar  Polygons  are  such  as  are  mutually  equi- 
angular, and  have  the  sides  about  the  equal  angles,  taken 
in  the  same  order,  proportional. 

17.  Homologous  Angles  in  similar  polygons  are  those 
which  are  equal  and  like  placed ;  and 

18.  The  Homologous  Sides  are  those  which  are  like  dis- 
posed about  the  homologous  angles. 

THEOREM   I. 

If  the  first  and  second  of  four  magnitudes  are  equal,  and 
also  the  third  and  fourth,  the  four  magnitudes  may  form  a 
proportion. 

Let  A,  B,  C,  and  D  represent  four  magnitudes,  such 
that  A  =  B  and  0  =  D ;  we  are  to  prove  that  A  :  B  : : 
0  :  D. 

Now,  by  hypothesis,  A  is  equal  to  B,  and  their  ratio  is 
therefore  1 ;  and  since,  by  hypothesis,  C  is  equal  to  D, 
their  ratio  is  also  1. 

Hence,  the  ratio  of  A  to  B  is  equal  to  that  of  C  to  D ; 
and,  (by  Def.  3), 

A  :  B  : :  0:1). 

Therefore,  four  magnitudes  which  are  equal,  two  and 
two,  constitute  a  proportion. 


BOOK  II.  63 


THEOREM    II. 


If  four  magnitudes  constitute  a  'proportions  the  product  of 
the  extremes  is  equal  to  the  product  of  the  means. 

Let  the  four  magnitudes  A,  B,  C,  and  D  form  the  pro- 
portion A  :  B  : :  0  :  B ;  we  are  to  prove  that  Ax  B 
=  Bx  C. 

The  ratio  of  A  to  B  is  expressed  by  -j  =  r. 
The  ratio  of  C  to  B  is  expressed  by  -^  =  r. 

Hence,  (Ax.  1),  -j  =  -. 

Multiplying  these  equals  each  by  A  x  C,  and  we  have 
Bx  C=AxB. 

Hence  the  theorem ;  if  four  magnitudes  are  in  propor- 
tion, etc. 

Cor.  1.  Conversely :  If  we  have  the  product  of  two  mag- 
nitudes equal  to  the  product  of  two  other  magnitudes,  they  will 
constitute  a  proportion  of  which  either  of  the  two  may  be  made 
the  extremes  and  the  other  two  the  means. 

Let  the  magnitudes  B  x  0=  A  x  B.    Dividing  both 
members  of  the   equation   by  A  x  C,   and  we   have 
B_B 
A~C 

Hence  the  proportion  A  :  B  : :  0  :  B. 

Cor.  2.  If  we  divide  both  members  of  the  equation 
Ax  B  =  Bx  C    by  .A, 

we  have         B  =  — -. — . 
A 

That  is,  to  find  the  fourth  term  of  a  proportion,  mul- 
tiply the  second  and  third  terms  together  and  divide  the  pro- 
duct by  the  first  term.  This  is  the  Rule  of  Three  of 
Arithmetic. 


64  GEOMETRY. 

This  equation  shows  that  any  one  of  the  four  terms 
can  be  found  by  a  like  process,  provided  the  other  three 
are  given. 

THEOREM   III. 

If  three  magnitudes  are  continued 'proportionals,  the  'product 
of  the  extremes  is  equal  to  the  square  of  the  mean. 

Let  A,  B,  and  0  represent  the  three  magnitudes : 

Then  A  :  B  : :  B  :  O,  (by  Def.  11). 

But,  (by  Th.  2),  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means ;  that  is,  A  x  0=  B\ 

Hence  the  theorem ;  if  three  magnitudes,  etc. 

THEOREM   IV. 

Equimultiples  of  any  two  magnitudes  have  the  same  ratio 
as  the  magnitudes  themselves  ;  and  the  magnitudes  and  their 
equimultiples  may  therefore  form  a  proportion. 

Let  A  and  B  represent  two  magnitudes,  and  mA  and 
mB  their  equimultiples. 

Then  we  are  to  prove  that  A  :  B  : :  mA  :  mB. 

The   ratio   of  A  to  B   is   —,  and  of  mA  to  mB  is 

A 

mB      B    .,  ,. 

— r  =  -r,  the  same  ratio. 
mA      A* 

Hence  the  theorem;  equimultiples  of  any  two  magni- 
tudes, etc. 

THEOREM    V. 

If  four  magnitudes  are  proportional,  they  will  he  propor- 
tional when  taken  inversely. 

If  A  :  B  : :  m A  :  mB,  then  B  :  A  : :  mB  :  m A ; 

For  in  either  case,  the  product  of  the  extremes  and 
means  are  manifestly  equal ;  or  the  ratio  of  the  couplets 
is  the  same. 

Hence  the  theorem ;  if  four  quantities  are  proportional, 
etc. 


BOOK   II.  65 


THEOREM    VI. 


Magnitudes  which  are  proportional  to  the  same  propor- 
tionals, are  proportional  to  each  other. 
If     A  :  B  =  P  :  Q  \     Then  we  are  to  prove  that 
and    a  :    b  =  P  :  Q)  A  :  B  =  a  :  b. 

13  Q 

From  the  1st  proportion,  —  =  ^ ; 

From  the  2d  "  -  =  ^5 

a      P 

Therefore,  by  (Ax.  1),      -j  =  -,  or  A  :  B  =  a  :  b. 

jA.       a 

Hence  the  theorem ;  magnitudes  which  are  proportional 
to  the  same  'proportionals,  etc. 

Cor.  1.  This  principle  may  be  extended  through  any 
number  of  proportionals. 

Cor.  2.  If  the  ratio  of  an  antecedent  and  consequent  of  one 
proportion  is  equal  to  the  ratio  of  an  antecedent  and  conse- 
quent of  another  proportion,  the  remaining  terms  of  the  two 
proportions  are  proportional. 

For,  if  A  :  B  : :  C  :  D 

and  M  :  N  ::  P:  Q 

in  Which  A=MthenC  =  P> 

hence  C  :  D  : :  P  :  Q. 

THEOREM   VII. 

If  any  number  of  magnitudes  are  proportional,  any  one  of 
the  antecedents  will  be  to  its  consequent  as  the  sum  of  all  the 
antecedents  is  to  the  sum  of  all  the  consequents. 

Let  A,  B,  C,  D,  U,  etc.,  represent  the  several  magni- 
tudes which  give  the  proportions 


A  :  B 
A  :  B 
A  :  B 
6* 


C:  D 
E  :  F 

Or  :  R,  etc.,  etc. 


66  GEOMETRY. 

To  which  we  may  annex  the  identical  proportion, 

A  :  B  : :  A  :  B. 
Now,  (by  Th.  2),  these  proportions  give  the  following 
equations, 

A  x  D  =  B  x   Q 
A  x  F  =  B  x  F 
A  x  H=  B  x   a 
A  x  B  =  B  x  A,  etc.  etc. 
From  which,  by  addition,  there  results  the  equation, 
A(B  +  D  +  F+H,  etc.)  =  B(A  +  0+  F  +  #,  etc.) 
But  the  sums  B  -f  J)  -f  F,  etc.,  and  A  +  C  +  F,  etc., 
may  be  separately  regarded  as  single  magnitudes ;  there- 
fore, (Th.  2), 

A  :  B  ::  A+C+F+  G,  etc.  :  B -f  D  +  F+  JT,  etc. 

Hence  the  theorem ;  if  any  number  of  magnitudes  are  pro- 
portional,  etc. 

THEOREM   VIII. 

If  four  magnitudes  constitute  a  proportion,  the  first  will  be 
to  the  sum  of  the  first  and  second  as  the  third  is  to  the  sum  of 
the  third  and  fourth. 

By  hypothesis,  A  :  B  : :  C:  D ;  then  we  are  to  prove 
that  A  :  A  +  B  ::  C  :  C+  D. 

By  the  given  proportion,  — -  =  -— . 

A         C 

Adding  unity  to  both  members,  and  reducing  them  to 
the  form  of  a  fraction,  we  have  — - —  =  — ^— .  Chang- 
ing this  equation  into  its  equivalent  proportional  form, 

we  have 

A  :  A  +  B  ::  O  :  C+  D. 

Hence  the  theorem  ;  if  four  magnitudes  constitute  a  pro- 
portion, etc. 

Cor.  If  we  subtract  each  member  of  the  equation  -j  = 


BOOK  II.  67 

-»  from  unity,  and  reduce  as  before,  we  shall  have 
A  :  A  —  B  ::  C  :  C—D. 

Hence  also ;  if  four  magnitudes  constitute  a  proportion, 
the  first  is  to  the  difference  between  the  first  and  second,  as  the 
third  is  to  the  difference  between  the  third  and  fourth. 

THEOREM   IX. 

If  four  magnitudes  are  proportional,  the  sum  of  the  first  and 
second  is  to  their  difference  as  the  sum  of  the  third  and  fourth 
is  to  their  difference. 

Let  A,  B,  C,  and  D  be  the  four  magnitudes  which  give 
the  proportion 

A  :  B  ::  0:1); 

we  are  then  to  prove  that  they  will  also  give  the  propor- 
tion 

A  +  B  :  A  —  B  ::  C+D  :  C  —  D. 
By  Th.  8  we  have  A  :  A  +  B  =  0  :  C+D. 
Also  by  Scholium,  same  Th.,  A  :  A  —  B  =  0  :  C —  D. 
Now,  if  we  change  the  order  of  the  means  in  these  pro- 
portions, which  may  be  done,  since  the  products  of  ex- 
tremes and  means  remain  the  same,  we  shall  have 

A  i  C  =  A  +  B  :  C+D. 
A  :  0  =  A  —  B  :  C—D. 
Hence,  (Th.  6),  we  have 

A  +  B  :  C+  D  =  A  —  B  :  O—B. 
Or,       A  +  B  :  A  —  B  =  C  +  I)  :  C—D. 

Hence  the  theorem ;  if  four  magnitudes  are  proportional, 
etc. 

THEOREM   X. 

If  four  magnitudes  are  proportional,  like  powers  or  like 
roots  of  the  same  magnitudes  are  also  proportional. 

If  the  four  magnitudes,  A,  B,  C,  and  D,  give  the  pro- 
portion 


68  GEOMETRY. 

A  :  B  : :  C  :  D, 
we  are  to  prove  that 

An  :  Bn  ::   (7"  :  Dw. 

7?        7) 

The  hypothesis  gives  the  equation  —  =  — .     Eaising 

A        0 

both  members  of  this  equation  to  the  nth.  power,  we  have 

Bn     Dn 

-j^  =  — ,  which,  expressed  in  its  equivalent  proportional 

form,  gives 

An  :  Bn  ::  Cn  :  Dn. 

If  n  is  a  wAete  number,  the  terms  of  the  given  propor- 
tion are  each  raised  to  a  power ;  but  if  n  is  a  fraction 
having  unity  for  its  numerator,  and  a  whole  number  for  its 
denominator,  like  roots  of  each  are  taken. 

As  the  terms  of  the  proportion  may  be  first  raised  to 
like  powers,  and  then  like  roots  of  the  resulting  propor- 
tion be  taken,  n  may  be  any  number  whatever. 

Hence  the  theorem ;  if  four  magnitudes,  etc. 

THEOREM   XI. 

If  four  magnitudes  are  proportional,  and  also  four  others, 
the  products  which  arise  from  multiplying  the  first  four  by  the 
second  four,  term  by  term,  are  also  proportional. 
Admitting  that  A  :     B  : :     0:1), 

and  X  :    Y  : :    M  :  N, 

We  are  to  show  that  AX :BY::  CM:  DN. 

B      J) 

From  the  first  proportion,  — .  ==  —  ; 

X     M' 

Multiply  these  equations,  member  by  member,  and 

BY^DN. 

AX      CM' 
Or,  AX  :  BY  ::  CM:  DN. 

The  same  would  be  true  in  any  number  of  proportions. 
Hence  the  theorem ;  if  four  magnitudes  are,  etc. 


From  the  second, 


BOOK   II.  69 

THEOREM    XII. 

If  four  magnitudes  are  proportional,  and  also  four  others, 
the  quotients  which  arise  from  dividing  the  first  four  by  the 
second  four,  term  by  term,  are  proportional. 

By  hypothesis,        A  :  B  : :  O  :  D, 

and  X  :  Y  ::  M :  1ST. 

Multiply  extremes  and  means,    AD  =  OB,       ( 1 ) 

and  XN=MY.      (2) 

Divide  (1)  by  (2),  and    ^  x  ^=  ^  x  |. 

Convert  these  four  factors,  which  make  two  equal  pro- 
ducts, into  a  proportion,  and  we  have 
A  .   B  _    C  9  D 
X1   Yi:  M:  N' 

By  comparing  this  with  the  given  proportions,  we  find 
it  is  composed  of  the  quotients  of  the  several  terms  of 
the  first  proportion,  divided  by  the  corresponding  terms 
of  the  second. 

Hence  the  theorem ;  if  four  magnitudes  are  proportional, 
etc. 

THEOREM    XIII. 

If  four  magnitudes  are  proportional,  we  may  multiply  the 
first  couplet,  the  second  couplet,  the  antecedents  or  the  conse- 
quents, or  divide  them  by  the  same  quantity,  and  the  results 
will  be  proportional  in  every  case. 

Let  the  four  magnitudes  A,  B,  0,  and  I)  give  the  pro- 
portion A  :  B  : :  0 :  D.  By  multiplying  the  extremes 
and  means  we  have 

A.D  =  B.O         (1) 

Multiply  both  members  of  this  equation  by  any  num- 
ber, as  a,  and  we  have 

aA.D  =  aB.O 

By  converting  this  equation  into  a  proportion  in  four 
different  ways,  as  follows : 


70 


GEOMETRY. 


aA  :  aB 
A  :  B  :  : 
a  A  :  B  : 
J.  :  a.B  : 


:   0  :  B 

aC  :  aD 
:  aO  :  B 
;  C  :  aB 


resuming  the  original  equation,  (1),  and  dividing  both 
members  by  a,  we  have 

A.B  _  B.O 
a  a 

This  equation  may  also  be  converted  into  a  proportion 
in  four  different  ways,  with  the  following  results : 


J 

x> 

0  :  B 

a 

a 

A  : 

B  :: 

0  .  D 
a  "    a 

A 

:  B  :: 

C1:I> 

a 

a 

A 

.  B  .. 

C:» 

a 

a 

Hence  the  theorem ;  if  four  magnitudes  are  in  proportion, 


etc. 


THEOREM    XIV. 


If  three  magnitudes  are  in  proportion,  the  first  is  to  the 
third  as  the  square  of  the  first  is  to  the  square  of  the  second. 

Let  A,  B,  and  C,  be  three  proportionals. 

Then  we  are  to  prove  that  A  :  C—A%\Bl 

By(Th.  3)  AC=B2 

Multiply  this  equation  by  the  numeral  value  of  A,  and 
we  have  A20=AB2 

This  equation  gives  the  following  proportion : 

A:  C=A2:B\ 

Hence  the  theorem. 

Remark.  —  It  is  now  proposed  to  make  an  application  of  the  pre- 
ceding abstract  principles  of  proportion,  in  geometrical  investigations 


BOOK   II. 


71 


THEOREM  XV. 
If  two  parallelograms  are  equal  in  area,  the  base  and  per- 
pendicular of  either  may  be  made  the  extremes  of  a  propor- 
tion, of  which  the  base  and  perpendicular  of  the  other  are  the 
means. 

Let  ABCD,     E  D  F c 

and  NLHM, 
be  two  paral- 
lelograms hav- 
ing equal  areas, 
by  hypothesis ;  then  we  are  to  prove  that 

AB  :  LN  : :  MK  :  BF, 
in  which  MK  and  BF  are  the 
altitudes  or  perpendiculars  of 
the  parallelograms. 

This  proportion  is  true,  if 
the  product  of  the  extremes 
is  equal  to  the  product  of  the  means ; 
that  is,  if  the  equation 

AB.BF  =  LN.MKiz  true. 
But  AB.BF  is  the  measure  of  the  rectangle  ABFE, 
(B.I.,  Th.  32,  Scholium),  and  this  rectangle  is  equal  in 
area  to  the  parallelegram  ABCD,  (B.  L,  Th.  27). 

In  the  same  manner,  we  may  prove  that  ~%N.MK  is 
the  measure  of  the  parallelogram  NLHM.  But  these 
two  parallelograms  have  equal  areas  by"  hypothesis. 

Therefore,  AB.BF  =  LN.MK  is  a  true  equation,  and 
(Th.  2,  Cor.  1),  gives  the  proportion 

AB  :  LN  : :  MK  :  BF. 
Hence  the  theorem ;  if  two  parallelograms  are  equal  in 
area,  etc. 

THEOREM  XVI. 

Parallelograms  having  equal  altitudes  are  to  each  other  as 
their  bases. 

Since  parallelograms  having  equal  bases  and  equal 
altitudes  are  equal  in  area,  however  much  their  angles 


72  GEOMETRY. 

may  differ,  we  can  suppose  the  two  parallelograms  under 
consideration  to  be  mutually  equiangular,  without  in  the 
least  impairing  the  generality  of  this  theorem.     There- 
fore,   let    ABOB 
and  AEFB  be  two  l — 7 — 7 — ; 

parallelograms         /     /     /    / 
having  equal  alti-       /     /     /     /    / 
tudes,and  let  them     /    /    /     /    /     /     /     / 
be    placed     with    A  B 

their  bases  on  the  same  line  AE,  and  let  the  side,  AB, 
be  common.  First  suppose  their  bases  commensurable, 
and  that  AE  being  divided  into  nine  equal  parts,  AB 
contains  four  of  those  parts. 

If,  through  the  points  of  division,  lines  be  drawn  paral- 
lel to  AB,  it  is  obvious  that  the  whole  figure,  or  the 
parallelogram,  AEFB,  will  be  divided  into  nine  equal 
parts,  and  that  the  parallelogram,  ABOB,  will  be  com- 
posed of  four  of  those  parts. 

Therefore,  ABOB  :  AEFB  : :  AB  :  AE  : :  4  :  9. 

Whatever  be  the  whole  numbers  having  to  each  other 
the  ratio  of  the  lines  AB  and  AE,  the  reasoning  would 
remain  the  same,  and  the  proportion  is  established  when 
the  bases  are  commensurable.  But  if  the  bases  are  not 
to  each  other  in  the  ratio  of  any  two  whole  numbers,  it 
remains  still  to  be  shown  that 

AEFB  :  ABOB  11  AE  1  AB    (1) 

If  this  propor- 
tion is  not  true, 
there  must  be  a 
line  greater  or  less 
than  AB,  to  which 

AE  will  have  the    A~  ~~b~l 

same  ratio  that  AEFB  has  to  ABOB. 

Suppose  the  fourth  proportional  greater  than  AB,  as 
AK,  then, 

AEFB  :  ABOB  ::  AE  :  AK    (2). 


C    M 


BOOK   II.  73 

If  we  now  divide  the  line  AE  into  equal  parts,  each 
less  than  the  line  BK,  one  point  of  division,  at  least,  will 
fall  between  B  and  K  Let  L  be  such  point,  and  draw 
LM  parallel  to  B  0. 

This  construction  makes  AE  and  AL  commensura- 
ble; and  by  what  has  been  already  demonstrated,  we 

have 

AEFD  :  ALMD  ::  AE  :  AL.     (3) 

Inverting  the  means  in  proportions  ( 2 )  and  ( 3 ),  they 

become 

AEFD  :  AE  : :  ABCD  :  AK; 

and  AEFD  :  AE  : :  ALMD  :  AL. 

Hence,  (Th.  6), 

ABCD  :  AK  : :  ALMD  :  AL. 

By  inverting  the  means  in  this  last  proportion,  we  have 

ABCD  :  ALMD  : :  AK  :  AL. 

But  AK  is,  by  hypothesis,  greater  than  AL;  hence,  if 
this  proportion  is  true,  ABCD  must  be  greater  than 
ALMD ;  but  on  the  contrary  it  is  less.  We  therefore 
conclude  that  the  supposition,  that  the  fourth  propor- 
tional, AK,  is  greater  than  AB,  from  which  alone  this 
absurd  proportion  results,  is  itself  absurd. 

In  a  similar  manner  it  can  be  proved  absurd  to  sup- 
pose the  fourth  proportional  less  than  AB. 

Therefore  the  fourth  term  of  the  proportion  ( 1 )  can  be 
neither  less  nor  greater  than  AB ;  it  is  then  AB  itself, 
and  parallelograms  having  equal  altitudes  are  to  each 
other  as  their  bases,  whether  these  bases  are  commensur- 
able or  not. 

Hence  the  theorem ;  Parallelograms  having  equal  bases, 
etc. 

Cor.  1.  Since  a  triangle  is  one  half  of  a  parallelogram 
having  the  same  base  as  the  triangle  and  an  equal  alti- 
tude, and  as  the  halves  of  magnitudes  have  the  same 
ratio  as  their  wholes ;  therefore, 
7 


74 


GEOMETRY. 


Triangles  having  the  same  or  equal  altitudes  are  to  each 
other  as  their  bases. 

Cor.  2.  Any  triangle  has  the  same  area  as  a  right- 
angled  triangle  having  the  same  base  and  an  equal  alti- 
tude; and- as  either  side  about  the  right  angle  of  aright- 
angled  triangle  may  be  taken  as  the  base,  it  follows  that 

Two  triangles  having  the  same  or  equal  bases  are  to  each 
other  as  their  altitudes. 

Cor.  3.  Since  either  side  of  a  parallelogram  may  be 
taken  as  its  base,  it  follows  from  this  theorem  that 

Parallelograms  having  equal  bases  are  to  each  other  as  their 
altitudes. 


THEOREM   XVII. 

If  lines  are  drawn  cutting  the  sides,  or  the  sides  produced,  of 
a  triangle  proportionally,  such  secant  lines  are  parallel  to  the 
base  of  the  triangle ;  and  conversely,  lines  drawn  parallel 
to  the  base  of  a  triangle  cut  the  sides,  or  the  sides  produced, 
proportionally. 

Let  ABC  be  any  triangle,  and 
draw  the  line  BE  dividing  the  sides 
AB  and  AC  into  parts  which  give 
the  proportion 

AD  :  DB  : :  AE  :  EC. 
"We  are  to  prove  that  BE  is  parallel 
\,oBC. 

If  BE  is  not  a  parallel  through 
the  point  B  to  the  line  BC,  suppose 
Bm  to  be  that  parallel ;  and  draw  the 
lines  BC  and  Bm. 

Now,  the  two  triangles  ABm  and 
mBC,  have  the  same  altitude,  since 
they  have  a  common  vertex,  B,  and  their  bases  in  the 
same  line,  A C;  hence,  they  are  to  each  other  as  their 
bases,  A m  and  mC,  (Th.  16,  Cor.  1). 


BOOK   II.  75 

That  is,      A  ADm  :  A  mDC  : :  Am  :  mC, 
Also,  A  Ami)  :  A  DmB  : :  AD  :  D#. 

But,  since  Dm  is  supposed  parallel  to  BC,  the  triangles 
DBm  and  D(7m  have  equal  areas,  because  they  are  on 
the  same  base  and  between  the  same  parallels,  (Th.  28, 
B.I). 

Therefore  the  terms  of  the  first  couplets  in  the  two 
preceding  proportions  are  equal  each  to  each,  and  conse- 
quently the  terms  of  the  second  couplets  are  also  propor- 
tional, (Th.  6). 

That  is,   AT)  :  DB  : :  Am  :  mC 

But  AD  :  DB  : :  AE  :  EC  by  hypothesis. 

Hence  we  again  have  two  proportions  having  the  first 
couplets,  the  same  in  both,  and  we  therefore  have 

AE  :  EG  ::  Am  :  mO 

By  alternation  this  becomes 

AE  :  Am  ::  EC  :  mO 

That  is,  AE  is  to  Am,  a  greater  magnitude  is  to  a  less, 
as  EC  is  to  mO,  a  less  to  a  greater,  which  is  absurd. 
Had  we  supposed  the  point  m  to  fall  between  E  and  0, 
our  conclusion  would  have  been  equally  absurd ;  hence 
the  suppositions  which  have  led  to  these  absurd  results 
are  themselves  absurd,  and  the  line  drawn  through  the 
point  D  parallel  to  BO  must  intersect  A 0  in  the  point 
E.  Therefore  the  parallel  and  the  line  BE  are  one  and 
the  same  line. 

Conversely :  If  BE  be  drawn  parallel  to  the  base  of  the 
triangle,  then  will 

AD  :  DB  : :  AE  :  EC 

For  as  before, 

A  ADE  :  a  EDO  ::  AE  :  EC 
and      A  DEB  :   A  AED  w  DB  \  AD 

Multiplying  the  corresponding  terms  of  these  propor- 


76  GEOMETRY. 

tions,  and  omitting  the  common  factor,  a  ADE,  in  the 
first  couplet,  we  have 

A  DEB  :  A  EDO  ::  AE  x  DB  :  EO  x  AD. 

But  the  a's  DEB  and  EDO  have  equal  areas,  (Th.  28, 
B.  I) ;  hence  AE  x  DB  =  EC  x  AD,  which  in  the  form 
of  a  proportion  is 

J.^  :  EG  : :  AD  :  DB 

or,  AD  :  DB  ::  AE  :  EC 

and  therefore  the  line  parallel  to  the  base  of  the  triangle, 
divides  the  sides  proportionally. 

It  is  evident  that  the  reasoning  would  remain  the  same, 
had  we  conceived  ADE  to  be  the  triangle  and  the  sides 
to  be  produced  to  the  points  B  and  0. 

Hence  the  theorem;  if  lines  are  drawn  cutting  the 
sides,  etc. 

Cor.  1.  Because  DE  is  parallel  to  BO,  and  intersects 
the  sides  AB  and  AC,  the  angles  ADE  and  ABO  are 
equal.  For  the  same  reason  the  angles  AED  and  A  OB 
are  equal,  and  the  A's  ADE  and  ABO  are  equiangular. 

Let  us  now  take  up  the  triangle  ADE,  and  place  it  on 
ABO;  the  angle  ADE  falling  on  [__  B,  the  side  AD  on 
the  side  AB,  and  the  side  DE  on  the  side  BO. 

Now,  since  the  angle  A  is  common,  and  the  angles 
AED  and  A  OB  are  equal,  the  side  AE  of  the  A  ADE, 
in  its  new  position,  will  be  parallel  to  the  side  A  0  of  the 
A  ABO. 

But  we  have  the  proportion 

AD  :  AE  ::  AB  :  AO 

Placing  the  angle  ADE  on  the  angle  ABO,  and  rea- 
soning as  before,  we  shall  have  the  proportion 
AD  :  DE  : :  AB  :  BO 

And  in  like  manner  it  may  be  shown  that 
AE  :  ED  ::  AC  :  OB 

That  is,  the  sides  about  the  equal  angles  of  equiangular 
triangles,  taken  in  the  same  order  yare  proportional,  and  the 
triangles  are  similar,  (Def.  16). 


BOOK   II 


77 


Cor.  2.  Two  triangles  having  an  angle  in  one  equal  to  an 
angle  in  the  other,  and  the  sides  about  these  equal  angles  pro- 
portional,  are  equiangular  and  similar. 

For,  if  the  smaller  triangle  be  placed  on  the  larger, 
the  equal  angles  of  the  triangles  coinciding,  then  will 
the  sides  opposite  these  angles  be  parallel,  and  the  trian- 
gles will  therefore  be  equiangular  and  similar. 


THEOREM    XVIII. 

If  any  triangle  have  its  sides  respectively  proportional  to 
the  like  or  homologous  sides  of  another  triangle,  each  to  each, 
then  the  two  triangles  will  be  equiangular  and  similar. 

Let  the  triangle  abc  have  its  sides  pro- 
portional to  the  triangle  ABO  ;  that  is,  ac 
to  A  0  as  cb  to  OB,  and  ac  to  A  0  as  ah  to 
AB ;  then  we  are  to  prove  that 
the  a's,  abc  and  ABO,  are  equi- 
angular and  similar. 

On  the  other  side  of  the  base, 
AB,   and  from  A,  conceive 
angle  BAB  to  be  drawn  =  to  the 

L   *r 

conceive 


and   from    the   point   B, 
the  angle  ABB  to   be 


drawn  =  to  the  [_  b.  Then  the  third  [__  B  must  be  = 
to  the  third  [_  c,  (B.  I,  Th.  12,  Cor.  2) ;  and  the  A  ABB 
will  be  equiangular  to  the  A  abc  by  construction. 

Therefore,  ac  :  ab  =  AB  :  AB 

By  hypothesis,  ac  :  ab  =  AO  :  AB 

Hence,  AB  :  AB  =  A  0  :  AB,  (Th.  6). 

In  this  last  proportion  the  consequents  are  equal; 
therefore,  the  antecedents  are  equal :  that  is, 
AB  =  AO 

In  the  same  manner  we  may  prove  that 

BB  =   OB 

7* 


78  GEOMETRY. 

But  AB  is  common  to  the  two  triangles";  therefore, 
the  three  sides  of  the  A  ABB  are  respectively  equal  to 
the  three  sides  of  the  A  ABC,  and  the  two  a's  are  equal, 
(B.  I,  Th.  21). 

But  the  A's  ABB,  and  abc,  are  equiangular  by  con- 
struction;  therefore,  the  A's,  ABO,  and  abc,  are  also 
equiangular  and  similar. 

Hence  the  theorem ;  if  any  triangle  have  its  sides,  etc, 

Second  Demonstration, 

Let  abc  and  ABC  be  two  triangles 
whose  sides  are  respectively  propor- 
tional, then  will  the  triangles  be  equi- 
angular and  similar. 

That  is,  [__a  =  l_A,  [_b  =  [_B,  and 
l_e=l_C. 

If  the  [__  c  be  in  fact 
equal  to  the  [_  C,  the  tri- 
angle abc  can  be  placed 
on  the  triangle  ABC,  ca 
taking  the  direction  of 
CA  and  cb  of  CB.  The 
line  ab  will  then  divide 
the  sides  CA  and  CB  proportionally,  and  will  therefore 
be  parallel  to  AB,  and  the  triangles  will  be  equiangular 
and  similar,  (Th.  17). 

But  if  the  L  c  be  not  equal  to  the  [__  C,  then  place  ac 
on  AC  as  before,  the  point  c  falling  on  C.  Under  the 
present  supposition  cb  will  not  fall  on  CB,  but  will  take 
another  direction,  CV,  on  one  side  or  the  other  of  CB. 
Make  CV  equal  to  cb  and  draw  aV. 

Now,  the  A  abc  is  represented  in  magnitude  and  posi- 
tion by  the  A  a  VC;  and  if,  through  the  point  a,  the  line 
ab  be  drawn  parallel  to  AB,  we  shall  have 

Ca  :   CA  ::  ab    :  AB; 
but  by  (Hy.)        Ca  :  CA  : :  aV  :  AB. 


BOOK   II 


79 


Hence,  (Th.  6), 

ab  :  AB  ::  aV  :  AB; 
which  requires  that  ab  =  a  V,  but  (Th.  22,  B.  I)  ab  can 
not  be  equal  to  a  V;  hence  the  last  proportion  is  absurd, 
and  the  supposition  that  the  [_  c  is  not  equal  to  the  [_  (7, 
which  leads  to  this  result,  is  also  absurd.  Therefore, 
the  [_  e  is  equal  to  the  [__  (7,  and  the  triangles  are  equi- 
angular and  similar. 

Hence  the  theorem ;  if  any  triangle  ham  its  sides,  etc. 


THEOREM    XIX. 

If  four  straight  lines  are  in  proportion,  the  rectangle  con- 
tained by  the  lines  which  constitute  the  'extremes,  is  equivalent 
to  that  contained  by  those  which  constitute  the  means  of  the 
proportion. 

Let  A,  B,  O,  D,  represent  the  four      A'        j 
lines;    then    we    are    to    show,   geo-        j         .  j 

metrically,  that  A  x  D  =  B  x  0.  Di i 


Place  A  and  B  at  right  angles  to  each 
other,  and  draw  the  hypotenuse.  Also  place 
0  and  D  at  right  angles  to  each  other,  and 
draw  the  hypotenuse.  Then  bring  the  two 
triangles  together,  so  that  0  shall  be  at  right 
angles  to  B,  as  represented  in  the  figure. 

Now,  these  two  A's  have  each  a  E.  [_, 
and  the  sides  about  the  equal  angles  are  pro- 
portional ;  that  is,  A  :  B  : :  0:1);  there- 
fore, (Th.  18),  the  two  A's  are  equiangular,  and  the 
acute  angles  which  meet  at  the  extremities  of  B  and  C, 
are  together  equal  to  one  right  angle,  and  the  lines  B 
and  0  are  so  placed  as  to  make  another  right  angle ; 
therefore,  also,  the  extremities  of  A,  B,  0,  and  Z>,  are  in 
one  right  line,  (Th.  3,  B.  I),  and  that  line  is  the  diag- 


\  B 

\ 

BC  ! 
C 

AD 

V 

80  GEOMETKY. 

onal  of  the  parallelogram  be.  By  Th.  31,  B.  I,  the 
complementary  parallelograms  about  this  diagonal  are 
equal ;  but,  one  of  these  parallelograms  is  B  in  length, 
and  Q  in  width,  and  the  other  is  D  in  length  and  A  in 
width;  therefore, 

B  x  0  =  A  x  D. 

Hence  the  theorem;  if  four  straight  lines  are  in  propor- 
tion, etc. 

Cor.  "When  B  =  Q,  then  A  x  D  =  B\  and  B  is  the 
mean  proportional  between  A  and  B.  That  is,  if  three 
straight  lines  are  in  proportion,  the  rectangle  contained 
by  the  first  and  third  lines  is  equivalent  to  the  square 
described  on  the  second  line. 

THEOREM    XX. 

Similar  triangles  are  to  one  another  as  the  squares  of  their 
homologous  sides. 

Let  ABC  and  DBF  be  two 
similar  triangles,  and  LQ  and 
MF  perpendiculars  to  the  sides 
AB  and  DE  respectively.  Then 
we  are  to  prove  that 
&ABQ:&BEF  =  AB*:BE\ 

By  the  similarity  of  the  tri- 
angles, we  have, 

AB    :  BE  =  LQ  :  MF 
But,         AB    :  DE  =  AB  :  BE 

Hence,     AB2  :  TW^~AB  x  LQ :  BE  x  MF. 

But,  (by  Th.  30,  B.  I),  AB  x  LQ  is  double  the  area 
of  the  A  ABC,  and  BE  x  MF  is  double  the  area  of  the 
A  BEF. 

Therefore,     aABC:ABEF::AB  x  LQ  :BExMF 

And,  (Th.  6),  A  ABQ:  A  DEF  =  AW  :  BE2. 

Hence  the  theorem ;  similar  triangles  are  to  one  another, 
etc. 


BOOK    II. 


81 


The  following  illustration  will  enable  the  learner  fully 
to  comprehend  this  important  theorem,  and  it  will  also 
serve  to  impress  it  upon  his  memory. 

Let  abc  and  ABO  represent  two  equiangular  triangles. 
Suppose  the  length  of 
the  side  ac  to  be  two 
units,  and  the  length 
of  the  corresponding 
side  A  0  to  be  three 
units. 

Eow,  drawing  lines 
through  the  points  of 

division  of  the  sides  ac  and  A  (7,  parallel  to  the  other  sides 
of  the  triangles,  we  see  that  the  smaller  triangle  is  com- 
posed of  four  equal  triangles,  while  the  larger  contains 
nine  such  triangles.     That  is, 

the  sides  of  the  triangles  are  as  2  :  3, 

and  their  areas  are  as  4  :  9  =  22  :  32. 


THEOREM   XXI. 


Similar  polygons  may  be  divided  into  the  same  number  of 
triangles;  and  to  each  triangle  in  one  of  the  polygons  there 
will  be  a  corresponding  triangle  in  the  other  polygon,  these 
triangles  being  similar  and  similarly  situated. 

Let  ABCDUsLnd  abcde 
be  two  similar  polygons. 
Now  it  is  obvious  that  we 
can  divide  each  polygon  E 
into  as  many  triangles  as 
the  figure  has  sides,  less 

two;  and  as  the  polygons  have  the  same  number  of  sides, 
the  diagonals  drawn  from  the  vertices  of  the  homologous 
angles  will  divide  them  into  the  same  number  of  tri- 
angles. 


82  GEOMETRY. 

Since  the  polygons  are  similar,  the  angles  EAB  and  eab, 
are  equal,  and 

EA  :  AB  ::  ea  :  ab. 

Hence  the  two  triangles,  EAB  and  eab,  having  an  angle 
in  the  one  equal  to  an  angle  in  the  other,  and  the  sides 
about  these  angles  proportional,  are  equiangular  and 
similar,  and  the  angles  ABE  and  abe  are  equal. 

But  the  angles  ABO  and  abc  are  equal,  because  the 
polygons  are  similar. 

Hence,  [_ABO—  [_ABE  =  [_abc  —  \_abe; 

that  is,  [_EBO<=[_ebc. 

The  triangles,  EAB  and  eab,  being  similar,  their  ho- 
mologous sides  give  the  proportion, 

AB  :  BE  ::  ab  :  be;         (1) 
and  since  the  polygons  are  similar,  the  sides  about  the 
equal  angles  B  and  b  are  proportional,  and  we  have 
AB  :  BO  : :  ab  :  be ; 

or,  BO  :  AB  ::  be  :  ab.       (2) 

Multiplying  proportions  (1)  and  (2),  term  by  term,  and 
omitting  in  the  result  the  factor  AB  common  to  the  terms 
of  the  first  couplet,  and  the  factor  ab  common  to  the 
terms  of  the  second,  we  have 

BO  :  BE  : :  be  :  be. 

Hence  the  A's  EBO and  ebe  are  equiangular  and  similar; 
and  thus  we  may  compare  all  of  the  triangles  of  one 
polygon  with  those  like  placed  in  the  other. 

Hence  the  theorem ;  similar  polygons  may  be  divided,  etc. 

THEOREM    XXII. 

The  perimeters  of  similar  polygons  are  to  one  another  as 
their  homologous  sides  ;  and  their  areas  are  to  one  another  as 
the  squares  of  their  homologous  sides. 

Let  ABODE  and  abode  be  two  similar  polygons ;  then 
we  are  to  prove  that  AB  is  to  the  sum  of  all  the  sides 


BOOK  II. 

83 

of  the  polygon  ABCB,  as 

B                  a             b 

ab  is  to  the  sum  of  all            / 

X^^. 

the  sides  of  the  polygon      y^^ 

abed.                                     E^^\^ 

-— ^C  e^^---        ^"^ 

¥e  have  the  identical               d 

d 

proportion 

AB  :  ab  : :  AB  : 

ab; 

and  since  the  polygons  are  similar,  we  may  write  the 

following : 

AB  :  ab  ::  BO  : 

be 

AB  :  ab  ::  OD  : 

ed 

AB  :  ab  : :  BE 

:  de,  etc.  etc. 

Hence,  (Th.  7), 
AB  :  ab  :  AB+BC+CB+BE,  etc.;  ab+bc+cd+de9  etc. 

Therefore,  the  perimeters  of  similar  polygons  are  to 
one  another  as  their  homologous  sides.  This  is  the  first 
part  of  the  theorem. 

Sinoe  the  polygons  are  similar,  the  triangles  BAB,  eab, 
are  similar,  and  if  the  triangle  BAB  is  a  part  expressed 

by  the  fraction  -,  of  the  polygon  to  which  it  belongs, 

71 

the  triangle  eab  is  a  like  part  of  the  other  polygon. 

Therefore,        EAB  i  eab  ::  ABCBEA  :  abedea. 

But,  (Th.  20)^ EAB  :  eab  : :  AB2  :  ab\ 

Therefore,  (Th.  6), 

ABCBEA  :  abedea  : :  AB2  :  ab\ 

Therefore,  the  similar  polygons  are  to  one  another  as 
the  squares  on  their  homologous  sides.  This  is  the 
second  part  of  the  theorem. 

Hence  the  theorem  ;  the  perimeters  of  similar  polygons 
are  to  one  another,  etc. 


THEOREM  XXIII. 

Two  triangles  which  have  an  angle  in  the  one  equal  to  an 
angle  in  the  other,  are  to  each  other  as  the  rectangle  of  the 
sides  about  the  equal  angles. 


84  GEOMETRY. 

Let  ABC  and  def  be  two  triangles  having  the  angles 
A  and  d  equal.  It  is  to 
be  proved  that  the  areas 
ABO  and  def  are  to  each 
other  as  AB.AO  is  to 
de.df. 

Conceive  the  triangle 
def  placed  on  the  tri- 
angle ABO,  so  that  d 
shall  fall  on  A,  and  de  on 
A B ;  then  df  will  fall  on 
AC,  because  the  [_'s  i 
and  d  are  equal.  On  AB,  lay  off  Ae,  equal  to  de ;  and 
on  AC,  lay  off  Af,  equal  to  df,  and  draw  ef  The  tri- 
angle Aef  will  then  be  equal  to  the  triangle  def.  Join 
B  and/. 

Now,  as  triangles  having  the  same  altitude  are  to  each 
other  as  their  bases,  (Th.  16,  Cor.  1),  we  have 

Aef    :  ABf   : :  Ae  :  AB 
also,  ABf  :  ABC  : :  Af:  AC 

Multiplying  these  proportions  together,  term  by  term, 
omitting  from  the  result  ABf  a  factor  common  to  the 
terms  of  the  first  couplet,  we  have 

Aef  :  ABC  ::  Ae  .  Af  :  AB  .  AC 

But  Aef  is  equal  to  def,  Ae  to  de,  and  Af  to  df;  therefore, 

def  :  ABC  : :  de  .  df  :  AB  .  AC 

Hence  the  theorem ;  two  triangles  which  have  an  angle,  etc. 

Scholium.  —  If  we  suppose  that 

AB  :  AC  ::  de  :  df 

the  two  triangles  will  be  similar ;  and  if  we  multiply  the  terms  of  the 
first  couplet  of  this  proportion  by  AC,  and  the  terms  of  the  second 
couplet  by  df  we  shall  have 

AB .AC :  AC*   : :  de  ^df  :  ^ 
or,  AB  .  AC  :  de  .  df  : :  AC2  :  df 


BOOK  II.  85 

Comparing  this  with  the  last  proportion  in  this  theorem,  and  we  have, 
(Th.6);  _       _ 

def:  ABC  ::  df  :  AC2 

Remark.  —  This  scholium  is  therefore  another  demonstration  of 
Theorem  20,  and  hence  that  theorem  need  not  necessarily  have  been 
made  a  distinct  proposition.  We  require  no  stronger  proof  of  the  cer- 
tainty of  geometrical  truth,  than  the  fact  that,  however  different  the 
processes  by  which  we  arrive  at  these  truths,  we  are  never  led  into 
inconsistencies  ;  but  whenever  our  conclusions  can  be  compared,  they 
are  found  to  harmonize  with  each  completely,  provided  our  premises 
are  true  and  our  reasoning  logical. 

It  is  hoped  that  the  student  will  lose  no  opportunity  to  exercise 
his  powers,  and  test  his  skill  and  knowledge,  in  seeking  original 
demonstrations  of  theorems,  and  in  deducing  consequences  and 
conclusions  from  those  already  established. 

THEOREM  XXIV. 

If  the  vertical  angle  of  a  triangle  be  bisected,  the  bisecting 
line  will  cut  the  base  into  segments  proportional  to  the  adja- 
cent sides  of  the  triangle. 

Let  ABO  be  any  triangle, 
and  the  vertical  angle,  0,  be  bi- 
sected by  the  straight  line  OD. 
Then  we  are  to  prove  that 

AD  :  DB  =  AC  :  OB. 

Produce  A  O  to  E,  making  A 
OB  =  OB,  and  draw  EB.  The  exterior  angle  A  OB,  of 
the  A  OEB,  is  equal  to  the  two  angles  E,  and  OBB; 
but  the  angle  E  =  OBE,  because  OB  =  OE,  and  the  tri- 
angle is  isosceles;  therefore  the  angle  AOD,  the  half  of 
the  angle  A  OB,  is  equal  to  the  angle  E,  and  BO  and  BE 
are  parallel,  (Cor.,  Th.  7,  B.  R 

Now,  as  ABE  is  a  triangle,  and  OB  is  parallel  to  BO,  ~ 
we  have  AD  :  DB  =  A  0  :  OE  or  OB,  (Th.  17). 

Hence  the  theorem ;  if  the  vertical  angle  of  a  triangle 
be  bisected,  etc. 
8 


86  GEOMETBY. 

THEOREM   XXV. 

If  from  the  right  angle  of  a  right-angled  triangle,  a  'per- 
pendicular is  drawn  to  the  hypotenuse  ; 

'  ft.  The  perpendicular  divides  the  triangle  into  two  similar 
triangles,  each  of  which  is  similar  to  the  whole  triangle. 

2.  The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

3.  The  segments  of  the  hypotenuse  are  in  proportion  to  the 
squares  on  the  adjacent  sides  of  the  triangle. 

4.  The  sum  of  the  squares  on  the  two  sides  is  equivalent  to 
the  square  on  the  hypotenuse. 

Let  BAO  be  a  triangle,  right  an- 
gled at  A ;  and  draw  AD  perpendicu- 
lar to  BO. 

1.  The  two  A's,  ABO  and  ABB,  B 
have  the  common  angle,  B,  and  the  right  angle  BAO  = 
the  right  angle  BDA;  therefore,  the  third  |__  0=  [__ 
BAB,  and  the  two  A's  are  equiangular,  and  similar. 
In  the  same  manner  we  prove  the  A  AB  0  similar  to 
the  A  ABO;  and  the  two  triangles,  ABB,  ABO,  being 
similar  to  the  same  A  ABO,  are  similar  to  each  other. 

2.  As  similar  triangles  have  the  sides  about  the  equal 
angles  proportional,  (Th.  17),  we  have 

BB  :  AB  ::  AB  :  OB; 

or,  the  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

3.  Again,  BO  :  BA  ::  BA  :  BB 
hence,  BA2  =  BO.BB        (1) 
also,  BOj_  OA  ::  OA  :  OD 
hence,  OA2  =  BO.OB         (2) 

Dividing  Eq.  (1)  by  Eq.  (2),  member  by  member,  we 
obtain 

~BA2        BB 

~OA2  "   OB 


BOOK   II.  87 

which,  in  the  form  of  a  proportion,  is 

~QJl    :~BA2  ::  CD  :  BD; 

that  is,  the  segments  of  the  hypotenuse  are  proportional  to  the 
squares  on  the  adjacent  sides. 
4.  By  the  addition  of  (1)  and  (2),  we  have 

SI2  +  CA*  m  BC(BD  +  CD)  =  BO2; 

that  is,  the  sum  of  the  squares  on  the  sides  about  the  right 
angle  is  equivalent  to  the  square  on  the  hypotenuse.  This  is 
another  demonstration  of  Theorem  39,  B.  I. 

Hence  the  theorem ;  if  from  the  right  angle  of  a  right- 
angled  triangle,  etc. 


88 


GEOMETKY. 


BOOK  III 


OF  THE  CIRCLE,  AND  THE  INVESTIGATION  OF  THEO- 
REMS DEPENDENT  ON  ITS  PROPERTIES. 

V,  DEFINITIONS. 


1.  *  A  Curved  Line  is  one  whose  consecutive  parts,  how- 
ever small,  do  not  lie  in  the  same  direction. 

2.  A  Circle  is  a  plane  figure  bounded  by  one  uniformly- 
curved  line,  all  of  the  points  of  which  are  at  the  same 
distance  from  a  certain  point  within,  called  the  center. 

3.  The  Circumference  of  a  cir- 
cle is  the  curved  line  that 
bounds  it. 

4.  The  Diameter  of  a  circle 
is  a  line  passing  through  the 
center,  and  terminating  at  both 
extremities  in  the  circumfer- 
ence. Thus,  in  the  figure,  0  is 
the  center  of  the  circle,  the 
curved  line  AGrBD  is  the  cir- 
cumference, and  AB  is  a  diameter. 

5.  The  Radius  of  a  circle  is  a  line  extending  from  the 
center  to  any  point  in  the  circumference.  Thus,  CD  is 
a  radius  of  the  circle. 

6.  An  Arc  of  a  circle  is  any  portion  of  the  circum- 
ference. 


*  The  first  six  of  the  above  definitions  have  been  before  given  among 
the  general  definitions  of  Geometry,  but  it  was  deemed  advisable  to 
reinsert  them  here. 


BOOK  III.  89 

7.  A  Chord  of  a  circle  is  the  line  connecting  the  ex- 
tremities of  an  arc. 

8.  A  Segment  of  a  circle  is  the  portion  of  the  circle  on 
either  side  of  a  chord. 

Thus,  in  the  last  figure,  ECrF  is  an  arc,  and  EF  is  a 
chord  of  the  circle,  and  the  spaces  bounded  by  the  chord 
EF,  and  the  two  arcs  EGrF  and  EDF,  into  which  it 
divides  the  circumference,  are  segments. 

9.  A  Tangent  to  a  circle  is  a  line  which,  meeting  the 
circumference  at  any  point,  will  not  cut  it  on  being 
produced.  The  point  in  which  the  tangent  meets  the 
circumference  is  called  the  point  of  tangency. 

10.  A  Secant  to  a  circle  is  a  line  which  meets  the  cir- 
cumference in  two  points,  and  lies  a  part  within  and  a 
part  without  the  circumference. 

11.  A  Sector  of  a  circle  is  a  portion  of  the  circle  included 
between  any  two  radii  and  their  intercepted  arc. 

Thus,  in  the  last  figure,  the  line  HL,  which  meets  the 
circumference  at  the  point  D,  but  does  not  cut  it,  is  a 
tangent,  D  being  the  point  of  tangency;  and  the  line 
MN,  which  meets  the  circumference  at  the  points  P  and 
Q,  and  lies  a  portion  within  and  a  portion  without  the 
circle,  is  a  secant.  The  area  bounded  by  the  arc  BJD,  and 
the  two  radii  OB,  CD,  is  a  sector  of  the  circle. 

12.  A  Circumscribed  Polygon  is 
one  all  of  whose  sides  are  tangent 
to  the  circumference  of  the  circle ; 
and  conversely,  the  circle  is  then 
said  to  be  inscribed  in  the  polygon. 

13.  An  Inscribed  Polygon  is  one 
the  vertices  of  whose  angles  are 
all  formed  in  the  circumference 
of  the  circle ;  and  conversely,  the  circle  is  then  said  to  be 
circumscribed  about  the  polygon. 

14.  A  Regular  Polygon  is  one  which  is  both  equiangu- 
lar and  equilateral. 

8* 


90 


GEOMETRY. 


The  last  three  definitions  are  illustrated  by  the  last 
figure. 

THEOREM    I. 

Any  radius  perpendicular  to  a  chord,  bisects  the  chord,  and 
also  the  arc  of  the  chord. 

Let  AB  be  a  chord,  0  the  center  of 
the  circle,  and  OE  a  radius  perpen- 
dicular to  AB ;  then  we  are  to  prove 
that  AB  t=4  BB,  and  A E  =  EB. 

Since  0  is  the  center  of  the  circle, 
AO=  BO,  OB  is  common  to  the  two 
A's  AOB  and  BOB,  and  the  angles 
at  B  are  right  angles;  therefore  the  two  A's  ABO  and 
BBO  are  equal,  and  AB  =  BB,  which  proves  the  first 
part  of  the  theorem. 

!N"ow,  as  AB  =  BB,  and  BB  is  common  to  the  two 
spaces,  ABB  and  BBE,  and  the  angles  at  B  are  right 
angles,  if  we  conceive  the  sector  OBE  turned  over  and 
placed  on  OAE,  OE  retaining  its  position,  the  point  B 
will  fall  on  the  point  A,  because  AB  =  BB  and  A  0  — 
BO;  then  the  arc  BE  will  fall  on  the  arc  AE\  otherwise 
there  would  be  points  in  one  or  the  other  arc  unequally 
distant  from  the  center,  which  is  impossible ;  therefore, 
the  arc  A E  =  the  arc  EB,  which  proves  the  second  part 
of  the  theorem. 

Hence  the  theorem. 

Oor.  The  center  of  the  circle,  the  middle  point  of 
the  chord  AB,  and  of  the  subtended  arc  AEB,  are 
three  points  in  the  same  straight  line  perpendicular  to 
the  chord  at  its  middle  point.  ISTow  as  but  one  perpen- 
dicular can  be  drawn  to  a  line  from  a  given  point  in  that 
line,  it  follows: 

1st.  That  the  radius  drawn  to  the  middle  point  of 
any  arc  bisects,  and  is  perpendicular  to,  the  chord  of 
the  arc. 


BOOK    III. 


91 


2d.  That  the  perpendicular  to  the  chord  at  its  middle 
point  passes  through  the  center  of  the  circle  and  the 
middle  of  the  subtended  arc. 


THEOREM    II. 

Equal  angles  at  the  center  of  a  circle  are  subtended  by 
equal  chords. 

Let  the  angle  A  OE  =  the  angle 
BEO;  then  the  two  isosceles  triangles, 
AOE,  and  EOB,  are  equal  in  all  re- 
spects, and  AE  —  EB. 

Hence  the  theorem. 


THEOREM    III. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  center. 

Let  AB  and  EF  be  equal  chords, 
and  0  the  center  of  the  circle.  From 
0,  draw  OG  and  OH,  perpendicular 
to  the  respective  chords.  These 
perpendiculars  will  bisect  the  chords, 
(Th.  1),  and  we  shall  have  A  G  =  EH. 
We  are  now  to  prove  that  OG  =  OH. 

Since  the  A's  EOH  and  AOG  are  right-angled,  we 
have,  (Th.  39,  B.  I), 


and, 


EH2  +  HO=~ECi 
~AO\ 


AG'  +  GO' 

By  subtracting  these  equations,  member  from  mem- 
ber, we  find  that 


EH1  —  AG1  -f  HO2  —  GO2  =  ~E02  —  AO2    (1) 
But  the  chords  are  equal  by  hypothesis,  hence  their 
halves,  EH  and  AG,  are  equal;  also  EO  =  AO,  being 
radii  of  the  circle.    "Wherefore, 


92  GEOMETRY. 


EE*  —  AG*  =  0 


and,  JSO  —  AOL  =  0. 

These  values  in  Equation  ( 1 )  reduce  it  to 

EX?  -GO_2=0 
or,  E02=G02 

and,  EO  =  GO. 

Hence  the  theorem. 

Oor.    Under  all  circumstances  we  have 

MS2  +  EQ2  =  AG2  +  GO2, 

because  the  sum  of  the  squares  in  either  member  of  the 
equation  is  equivalent  to  the  square  of  the  radius  of  the 
circle. 

ISTow,  if  we  suppose  HQ  greater  than  GO,  then  will 
HO2  be  greater  than  GO2.  Let  the  difference  of  these 
squares  be  represented  by  d. 

Subtracting  GO2  from  both  members  of  the  above 
equation,  we  have 

EE2+d  =  AG2 


whence,        A  G2  >  JEM2,  and  A  G  >  HE. 

Therefore,  AB,  the  double  of  AG,  is  greater  than  EF, 
the  double  of  UE;  that  is,  of  two  chords  in  the  same  or 
equal  circles,  the  one  nearer  the  center  is  the  greater. 

The  equation,  ME2  +  EO2  =  AG2  +  ~G02,  being  true, 
whatever  be  the  position  of  the  chords,  we  may  suppose 
GO  to  have  any  value  between  0  and  A 0,  the  radius  of 
the  circle. 

When  GO  becomes  zero,  the  equation  reduces  to 
EE2+~E02  =  AG2  =  B*; 
that  is,  under  this  supposition,  AG  coincides  with  AO, 
and  AB  becomes  the  diameter  of  the  circle,  the  greatest 
chord  that  can  be  drawn  in  it. 


BOOK    III.  93 


THEOREM    IV. 


A  line  tangent  to  the  circumference  of  a  circle  is  at  right 
angles  with  the  radius  drawn  to  the  point  of  contact. 

Let  A  0  be  a  line  tangent  to  the  circle 
at  the  point  B,  and  draw  the  radius,  EB, 
and  the  lines,  AE  and  CE. 

Now,  we  are  to  prove  that  EB  is  per- 
pendicular to  AC.    Because  B  is  the 
only  point  in  the  line  AC  which  meets 
the  circle,  (Def.  9,  B.  II),  any  other  line, 
as  AE  or  CE,  must  be  greater  than  EB; 
therefore,  EB  is  the  shortest  line  that  can  be  drawn  from 
the  point  E  to  the  line  AG;  and  EB  is  the  perpendicu- 
lar to  AC,  (Th.  23,  B.I). 

Hence  the  theorem. 

THEOREM   V. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  subtend 
or  stand  on  equal  portions  of  the  circumference. 

Conceive  two  equal  circles,  and  two  equal  chords  drawn 
within  them.  Then,  conceive  one  circle  taken  up  and 
placed  upon  the  other,  center  upon  center,  in  such  a  po- 
sition that  the  two  equal  chords  will  fall  on,  and  exactly 
coincide  with,  each  other;  the  circles  must  also  coin- 
cide, because  they  are  equal ;  and  the  two  arcs  of  the  two 
circles  on  either  side  of  the  equal  chords  must  also  coin- 
cide, or  the  circles  could  not  coincide ;  and  magnitudes 
which  coincide,  or  exactly  fill  the  same  space,  are  in  all 
respects  equal,  (Ax.  10). 

Hence  the  theorem. 


94 


GEOMETRY. 


THEOREM    VI, 


Through  three  given  points,  not  in  the  same  straight  line, 
one  circumference  can  be  made  to  pass,  and  but  one. 

Let  A,  B,  and  0  be  three  given 

points,  not  in  the  same  straight 

line,  and  draw  the  lines  AB  and 

BO.     If  a  circumference  is  made 

to  pass  through  the  two  points  A 

and  B,  the  line  AB  will  be  a  chord 

to  such  a  circle ;  and  if  a  chord  is 

bisected  by  a  line  at  right  angles, 

the  bisecting  line  will  pass  through 

the  center  of  the  circle,  (Cor.,  Th.  1) ;  therefore,  if  we 

bisect  the  line  AB,  and  draw  DF,  perpendicular  to  N, 

at  the  point  of  bisection,  any  circumference  that  can 

pass  through  the  points,  A  and  B,  must  have  its  center 
somewhere  in  the  line  DF.  And  if  we  draw  EGr  at 
right  angles  to  BO  at  its  middle  point,  any  circumference 
that  can  pass  through  the  points  B  and  0  must  have  its 
center  somewhere  in  the  line  EG.  JS"ow,  if  the  two  lines, 
DF  and  ECr,  meet  in  a  common  point,  that  point  will  be 
a  center,  about  which  a  circumference  can  be  drawn  to 
pass  through  the  three  points,  A,  B,  and  0,  and  DF  and 
EG  will  meet  in  every  case,  unless  they  are  parallel ;  but 
they  are  not  parallel,  for  if  they  were,  it  would  follow 
(Th.  5,  B.  I)  that,  since  DF  is  intersected  at  right  angles 
by  the  line  AB,  it  must  also  be  intersected  at  right  angles 
by  the  line  BO,  having  a  direction  different  from  that  of 
AB ;  which  is  impossible,  (Th.  7,  B.  I). 

Therefore  the  two  lines  will  meet ;  and,  with  the  point 
H,  at  which  they  meet,  as  a  center,  and  HB  —  HA  =  SO 
as  a  radius,  one  circumference,  and  but  one,  can  be  made 
to  pass  through  the  three  given  points. 
Hence  the  theorem. 


BOOK   III.  95 

THEOREM  VII. 

If  two  circles  touch  each  other,  either  internally  or  exter- 
nally, the  two  centers  and  the  point  of  contact  will  he  in  one 
right  line. 

Let  two  circles  touch  each 
other  internally,  as  represented 
at  A,  and  conceive  AB  to  be  a 
tangent  at  the  common  point  A. 
Now,  if  a  line,  perpendicular  to 
AB,  be  drawn  from  the  point 
A,  it  must  pass  through  the 
center  of  each  circle,  (Th.  4) ; 

and  as  but  one  perpendicular  can  be  drawn  to  a  line  at  a 
given  point  in  it,  A,  C,  and  B,  the  point  of  contact  and 
the  two  centers  must  be  in  one  and  the  same  line. 

Next,  let  two  circles  touch  each  other  externally,  and 
from  the  point  of  contact  conceive  the  common  tangent, 
AB,  to  be  drawn. 

Then  a  line,  AG,  perpendicular  to  AB,  will  pass 
through  the  center  of  one  circle,  (Th.  4),  and  a  per- 
pendicular, AB,  from  the  same  point,  A,  will  pass 
through  the  center  of  the  other  circle ;  hence,  BAO  and 
BAB  are  together  equal  to  two  right  angles ;  therefore 
CAB  is  one  continued  straight  line,  (Th.  3,  B.  I). 

Cor.  "When  two  circles  touch  each  other  internally,  the 
distance  between  their  centers  is  equal  to  the  difference 
of  their  radii ;  and  when  they  touch  each  other  extern- 
ally, the  distance  between  their  centers  is  equal  to  the 
sum  of  their  radii. 

THEOREM   VIII. 

An  angle  at  the  circumference  of  any  circle  is  measured  by 
one  half  the  arc  on  which  it  stands. 

In  this  work  it  is  taken  as  an  axiom  that  any  angle 
whose  vertex  is  at  the  center  of  a  circle,  is  measured  by 


96 


GEOMETRY. 


the  arc  on  which  it  stands ;  and  we  now  proceed  to  prove 
that  when  the  arcs  are  equal,  the  angle  at  the  circumference 
is  equal  to  one  half  the  angle  at  the  center. 

Let  A  OB  be  an  angle  at  the  center, 
and  D  an  angle  at  the  circumference, 
and  at  first  suppose  D  in  a  line  with 
A  0.  We  are  now  to  prove  that  the 
angle  A  OB  is  double  the  angle  D. 

The  A  DCB  is  an  isosceles  triangle, 
because  OB  =  OB ;  and  its  exterior 
angle,  A  OB,  is  equal  to  the  two  interior  angles,  B,  and 
OBB,  (Th.  12,  B.  I),  and  since  these  two  angles  are  equal 
to  each  other,  the  angle  AOB  is  double  the  angle  at 
B.  But  A  OB  is  measured  by  the  arc  AB ;  therefore  the 
angle  B  is  measured  by  one  half  the  arc  AB. 

Next,  suppose  B  not  in  a  line  with 
A  0,  but  at  any  point  in  the  circum- 
ference, except  on  AB ;  produce  BO 
toE. 

Now,  by  the  first  part  of  this 
theorem, 

the  angle  EOB  =  2EBB, 

also,  BOA  =  2EBA, 

by  subtraction,  AOB  =  2 ABB. 

But  AOB  is  measured  by  the  arc  AB;  therefore  ABB 
or  the  angle  D,  is  measured  by  one  half  of  the  same  arc. 
Hence  the  theorem. 


THEOREM   IX. 

An  angle  in  a  semicircle  is  a  right  angle ;  an  angle  in  a 
segment  greater  than  a  semicircle  is  less  than  a  right  angle  ; 
and  an  angle  in  a  segment  less  than  a  semicircle  is  greater 
than  a  right  angle. 

If  the  angle  AOB  is  in  a  semicircle,  the  opposite  seg- 
ment, ABB,  on  which  it  stands,  is  also  a  semicircle ;  and 
the  angle  AOB  is  measured  by  one  half  the  arc  ABB. 


BOOK    III. 


97 


(Th.  8) ;  that  is,  one  half  of  180°,  or  90°,  which  is  the 
measure  of  a  right  angle. 

If  the  angle  ACB  is  in  a  segment 
greater  than  a  semicircle,  then  the 
opposite  segment  is  less  than  a  semi- 
circle, and  the  measure  of  the  angle 
is  less  than  one  half  of  180°,  or  less 
than  a  right   angle.     If  the   angle 
ACB  is   in  a  segment  less  than  a 
semicircle,  then  the  opposite  segment,  ABB,  on  which 
the  angle  stands,  is  greater  than  a  semicircle,  and  its  half 
is   greater  than  90° ;    and,  consequently,  the   angle   is 
greater  than  a  right  angle. 

Hence  the  theorem. 

Cor.  Angles  at  the  circumference, 
and  standing  on  the  same  arc  of  a 
circle,  are  equal  to  one  another ;  for 
all  angles,  as  BAC,  BBC,  BBC,  are 
equal,  because  each  is  measured  by 
one  half  of  the  arc  BC.  Also,  if  the 
angle  BBC  is  equal  to  CEG-,  then 
the  arcs  BC  and  CG-  are  equal,  be- 
cause their  halves  are  the  measures  of  equal  angles. 

THEOREM    X. 

The  sum  of  two  opposite  angles  of  any  quadrilateral  in- 
scribed in  a  circle,  is  equal  to  two  right  angles. 

Let  ACBD  represent  any  quadri- 
lateral inscribed  in  a  circle.  The 
angle  ACB  has  for  its  measure,  one 
half  of  the  arc  ABB,  and  the  angle 
ABB  has  for  its  measure,  one  half  of 
the  arc  ACB;  therefore,  by  addition, 
the  sum  of  the  two  opposite  angles  at 
C  and  B,  are  together  measured  by 
one  half  of  the  whole  circumference,  or  by  180  degrees, 
=  two  right  angles.  Hence  the  theorem. 
9 


98 


GEOMETRY. 


THEOREM  XI. 

An  angle  formed  by  a  tangent  and  a  chord  is  measured  by 

one  half  of  the  intercepted  arc. 

Let  AB  be  a  tangent,  and  AD  a 
chord,  and  A  the  point  of  contact ; 
then  we  are  to  prove  that  the  angle 
BAD  is  measured  by  one  half  of  the 
arc  AED. 

From  A  draw  the  radius  A  C;  and 
from  the  center,  0,  draw  CE  per- 
pendicular to  AD. 

The    l_BAD  +  [_DAO=  90°,  (Th.  4). 

Also,        ]^C+l_DAC=  90°,  (Cor.  4,  Th.  12,  B.  I). 

Therefore,  by  subtraction,  BAD  —  (7=0; 

by  transposition,  the  angle  BAD  =  0. 

But  the  angle  0,  at  the  center  of  the  circle,  is  measured 
by  the  arc  AE,  the  half  of  AED ;  therefore,  the  equal 
angle,  BAD,  is  also  measured  by  the  arc  AE,  the  half 
of  AED. 

Hence  the  theorem. 

See  Th.  13,  for  another  proof. 


THEOREM   XII. 

An  angle  formed  by  a  tangent  and  a  chord,  is  equal  to  an 
angle  in  the  opposite  segment  of  the  circle. 

Let  AB  be  a  tangent,  and  AD  a 
chord,  and  from  the  point  of  contact, 
A,  draw  any  angles,  as  AOD,  and 
AED,  in  the  segments.  Then  we  are 
to  prove  that  [__  BAD  =[__ACD,  and 
[_  aAD  =  L  AED. 

By  Th.  11,  the  angle  BAD  is  meas- 
ured by  one  half  the  arc  AED ;  and 
as  the  angle  ACD  is  measured  by  one  half  of  the  same 
arc,  (Th.  8),  we  have  [_  BAD  =  [_ACD. 


BOOK   III.  99 

Again,  as  AEBO  is  a  quadrilateral,  inscribed  in  a 
circle,  the  sum  of  the  opposite  angles, 

AOB  +  AEB  =  2  right  angles.     (Th.  10). 

Also,  the  sum  of  the  angles 

BAB  +  BAG  =  2  right  angles.     (Th.  1,  B.  I). 

By  subtraction  (and  observing  that  BAB  has  just  been 
proved  equal  to  AOB),  we  have, 

AEB  — BAG  =  0. 
Or,  by  transposition,         AEB  =  BAG. 

Hence  the  theorem. 

THEOREM    XIII. 

Arcs  of  the  circumference  of  a  circle  intercepted  by  paral- 
lel chords,  or  by  a  tangent  and  a  parallel  chord,  are  equal. 

Let  AB  and  OB  be  parallel  chords, 
and  draw  the  diagonal,  AB ;  now,  be- 
cause AB  and  OB  are  parallel,  the 
angle  BAB  =  the  angle  ABO  (Th.  6,  B. 
I) ;  but  the  angle  BAB  has  for  its  meas- 
ure, one  half  of  the  arc  BB;  and  the 
angle  ABO  has  for  its  measure,  one  half  of  the  arc  A 0, 
(Th.  8) ;  and  because  the  angles  are  equal,  the  arcs  are 
equal ;  that  is,  the  arc  BB  =  the  arc  AO. 

Next,  let  EF  be  a  tangent,  parallel  to  a  chord,  OB,  and 
from  the  point  of  contact,  G,  draw  GB. 

Since  EF  and  OB  are  parallel,  the  angle  OBG  =  the 
angle  BGF.  But  the  angle  OBG  has  for  its  measure, 
one-half  of  the  arc  OG,  (Th.  8) ;  and  the  angle  BGF 
has  for  its  measure,  one  half  of  the  arc  GB,  (Th.  11) ; 
therefore,  these  equal  measures  of  equals  must  be  equal ; 
that  is,  the  arc  OG  =  the  arc  GB. 

Hence  the  theorem. 


100 


GEOMETRY. 


THEOREM   XIV. 


When  two  chords  intersect  each  other  within  a  circle,  the 
angle  thus  formed  is  measured  by  one  half  the  sum  of  the  two 
intercepted  arcs. 

Let  AB  and  CD  intersect  each 
other  within  the  circle,  forming  the 
two  angles,  E  and  E',  with  their 
equal  vertical  angles. 

Then,  we  are  to  prove  that  the 
angle  E  is  measured  by  one  half  the 
sum  of  the  arcs  A  0  and  BD;  and 
the  angle  E1  is  measured  by  one  half  the  sum  of  the 
arcs  AB  and  OB. 

First,  draw  AF  parallel  to  CD,  and  FD  will  be  equal 
to  AC,  (Th.  13);  then,  by  reason  of  the  parallels,  |__  BAF 
=  |_  E.  But  the  angle  BAF  is  measured  by  one  half 
of  the  arc  BDF;  that  is,  one  half  of  the  arc  BD  plus  one 
half  of  the  arc  AC 

Now,  as  the  sum  of  the  angles  B  and  E'  is  equal  to 
two  right  angles,  that  sum  is  measured  by  one  half  the 
whole  circumference. 

But  the  angle  E,  alone,  as  we  have  just  proved,  is 
measured  by  one  half  the  sum  of  the  arcs  BD  and  AC; 
therefore,  the  other  angle,  E1,  is  measured  by  one  half 
the  sum  of  the  other  parts  of  the  circumference, 

AD  +  OB. 

Hence  the  theorem. 


THEOREM    XV, 


When  two  secants  intersect,  or  meet  each  other  without  a 
circle,  the  angle  thus  formed  is  measured  by  one  half  the  dif- 
ference of  the  intercepted  arcs. 


BOOK    III. 


101 


Let  BE  and  BE  be  two  secants 
meeting  at  E ;  and  draw  A F  parallel  to 
OB.  Then,  by  reason  of  the  parallels, 
the  angle  E,  made  by  the  intersection 
of  the  two  secants,  is  equal  to  the 
angle  BAF.  But  the  angle  BAF  is 
measured  by  one  half  the  arc  BF; 
that  is,  by  one  half  the  difference  be- 
tween the  arcs  BB  and  AC. 

Hence  the  theorem. 


THEOREM   XVI. 


The  angle  formed  by  a  secant  and  a  tangent  is  measured 
by  one  half  the  difference  of  the  intercepted  arc. 

Let  BQ  be  sl  secant,  and  OB  a  tan- 
gent, meeting  at  0.  We  are  to  prove 
that  the  angle  formed  at  0,  is  meas- 
ured by  one  half  the  difference  of  the 
arcs  BB  and  BA. 

From  A,  draw  AE  parallel  to  OB ; 
then  the  arc  AB  =  the  arc  BE; 
BB  —  BE  =  BE;  and  the  [_BAE  = 
L  0.  But  the  angle  B A E  is  measured 
by  one  half  the  arc  BE,  (Th.  8,)  that  is,  by  one  half 
the  difference  between  the  arcs  BB  and  AB;  there- 
fore, the  equal  angle,  0,  is  measured  by  one  half  the 
arc  BE. 

Hence  the  theorem. 


THEOREM   XVII. 

When  two  chords  intersect  each  other  in  a  circle,  the  rect- 
angle contained  by  the  segments  of  the  one,  will  be  equivalent 
to  the  rectangle  contained  by  the  segments  of  the  other. 
9* 


102 


GEOMETRY. 


Let  AB  and  CD  be  two  chords  inter- 
secting each  other  in  E.  Then  we  are 
to  prove  that  the  rectangle  AE  x  EB  = 
the  rectangle  OE  x  ED. 

Draw  the  lines  AD  and  CB,  forming 
the  two  triangles  AED  and  CEB.  The 
angles  B  and  D  are  equal,  because  they 
are  each  measured  by  one  half  the  arc,  AC.  Also  the 
angles  A  and  0  are  equal,  because  each  is  measured  by 
one  half  the  arc,  DB ;  and  L  AED  =  [_  CEB,  because 
they  are  vertical  angles ;  hence,  the  triangles,  AED  and 
CEB,  are  equiangular  and  similar.  But  equiangular  tri- 
angles have  their  sides  about  the  equal  angles  propor- 
tional, (Cor.  1,  Th.  17,  B.  II);  therefore,  AE  and  ED, 
about  the  angle  E,  are  proportional  to  CE  and  EB,  about 
the  same  or  equal  angle. 

That  is,  AE  :  ED  : :  CE  :  EB; 

Or,  (Th.  19,  B.  n),     AExEB=  CEx  ED. 

Hence  the  theorem. 

Cor.  When  one  chord  is  a  diameter,  and  the  other  at  right 
angles  to  it,  the  rectangle  contained  by  the  segments  of  the 
diameter  is  equal  to  the  square  of  one  half  the  other  chord; 
or  one  half  of  the  bisected  chord  is  a  mean  proportional  be- 
tween the  segments  of  the  diameter. 

For,  ADxDB*=FD  x  DE.  But,  if 
AB  passes  through  the  center,  C,  at 
right  angles  to  FE,  then  FD  =  DE 
(Th.  1) ;  and  in  the  place  of  FD,  write 
its  equal,  DE,  in  the  last  equation,  and 
we  have 

ADxDB  =  DE2, 

or,  (Th.  3,  B.  IT),    AD  :  DE  : :  DE  :  DB. 

Put,  DE  =x,  CD  =  y,  and  CE  =  R,  the  radius  of  the 
circle. 


BOOK    III. 


103 


Then  AD  =  B—y,  and  DB  =  B  +  y.  With  this  nota- 
tion, 

AD  x  DB  =  DE2 

becomes,  (B — y)  (B  +  y)  —  x2 

or,  B2  —  y2  =  x2 

or,  B2  =  x2+y2 

That  is,  the  square  of  the  hypotenuse  of  the  right-angled 
triangle,  DOE,  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides. 


THEOREM   XVIII. 

If  from  a  point  without  a  circle,  a  tangent  line  be  drawn  to 
the  circumference,  and  also  any  secant  line  terminating  in  the 
concave  arc,  the  square  of  the  tangent  will  be  equivalent  to  the 
rectangle  contained  by  the  whole  secant  and  its  external  seg- 
ment. 

Let  A  be  a  point  without  the 
circle  DEGr,  and  let  AD  be  a 
tangent  and  AE  any  secant  line. 

Then  we  are  to  prove  that 
AOxAE^AD2. 
In  the  two  triangles,  ADE  and 
ADC,  the  angles  ADO  and  AED 
are  equal,  since  each  is   meas- 
ured by  one  half   of  the  same 
arc,  DO;   the  angle  A  is  com-, 
mon  to  the  two  triangles ;  their 

third  angles  are  therefore  equal,  and  the  triaugles  are 
equiangular  and  similar. 

Their  homologous  sides  give  the  proportion 
AE  :  AD  : :  AD  :  AO 

whence,  AE  x  AO=  AD2 

Hence  the  theorem. 

Oor.   If  AE  and  AF  are  two  secant  lines  drawn  from 
the  same  point  without  the  circumference,  we  shall  have 


104  GEO  ME  THY. 

ACx  AE=AD2 
and,  ABxAF=AD2 

hence,  AC  x  AE  =  AB  X  AF, 

which,  in  the  form  of  a  proportion,  gives 
AC  :  AF  ::AB  :  AE. 

That  is,  ^6  secants  are  reciprocally  proportional  to  their  ex- 
ternal segments. 

Scholium.  —  By  means  of  this  theorem  we  can  determine  the  diam- 
eter of  a  circle,  when  we  know  the  length  of  a  tangent  drawn  from  a 
point  without,  and  the  external  segment  of  the  secant,  which,  drawn 
from  the  same  point,  passes  through  the  center  of  the  circle. 

Let  Am  be  a  secant  passing  through  the  center,  and 
suppose  the  tangent  AD  to  be  20,  and  the  external  seg- 
ment, An,  of  the  secant  to  be  2.  Then,  if  D  denote  the 
diameter,  we  shall  have 

whence,  Am  x  An  -  2  (2  +  D)  =  4  +  2D  =  (20)2  =  400, 
22)  =396,  and  2)  =  198. 

If  An,  the  height  of  a  mountain  on  the  earth,  and  AD, 
the  distance  of  the  visible  sea  horizon,  be  given,  we  may 
determine  the  diameter  of  the  earth. 

For  example ;  the  perpendicular  height  of  a  mountain 
on  the  island  of  Teneriffe  is  about  3  miles,  and  its  summit 
can  be  seen  from  ships  when  they  are  known  to  be  154 
or  155  miles  distant ;  what  then  is  the  diameter  of  the 
earth  ? 

Designate,  as  before,  the  diameter  by  2>.  Then  Am  = 
3  +  2),  and  Am  x  An  =  9  +  32).  AD  =  154,  5 ;  hence, 
9  +  32)  =  (154,  5)2  =  23870.  25,  from  which  we  find  D  = 
7953.T3,  which  differs  but  little  from  the  true  diameter 
of  the  earth. 

One  source  of  error,  in  this  mode  of  computing  the 
diameter  of  the  earth,  is  atmospheric  refraction,  the  ex- 
planation of  which  does  not  belong  here. 


BOOK  III.  105 

THEOREM   XIX. 

If  a  circle  he  described  about  a  triangle,  the  rectangle  con- 
tained by  two  sides  of  the  triangle  is  equivalent  to  the  rectangle 
contained  by  the  perpendicular  let  fall  on  the  third  side,  and 
the  diameter  of  the  circumscribing  circle. 

Let  ABO  be  a  triangle,  AO  and 
OB,  the  sides,  OB  the  perpendicular 
let  fall  on  the  base  AB,  and  OB  the 
diameter  of  the  circumscribing  circle. 
Then  we  are  to  prove  that 

AOx  OB=  OEx  OB. 

The  two  A's,  AOB  and  OEB,  are 
equiangular,  because  [_A—[__B,  both 
being  measured  by  the  half  of  the  arc  OB;  also,  ABO  is 
a  right  angle,  and  is  equal  to  OBB,  an  angle  in  a  semi- 
circle, and  therefore  a  right  angle ;  hence,  the  third  angle, 
AOB  =  \_BOE,  (Th.  12,  Cor.  2,  B.  I).  Therefore,  (Cor., 
Th.  17,  B.  II), 

AO  :  OB  ::  OB  :  OB 

and,  AOx  BO=  OB x  OB. 

Hence  the  theorem ;  if  a  circle,  etc. 

Oor.  The  continued  product  of  three  sides  of  a  triangle  is 
equal  to  twice  the  area  of  the  triangle  into  the  diameter  of  its 
circumscribing  circle. 

Multiplying  both  members  of  the  last  equation  by  AB, 
and  we  have, 

AO  x  BO  x  AB  =  OB  x  {AB  x  OB). 

But  OB  is  the  diameter  of  the  circle,  and  (AB  x  OB) 
=  twice  the  area  of  the  triangle ; 

Therefore,  AO  X  OB  x  AB  =  diameter  multiplied 
by  twice  the  area  of  the  triangle. 


105  GEOMETRY. 

THEOREM   XX. 

The  square  of  a  line  bisecting  any  angle  of  a  triangle,  to* 
gether  with  the  rectangle  of  the  segments  into  which  it  cuts  the 
opposite  side,  is  equivalent  to  the  rectangle  of  the  two  sides 
including  the  bisected  angle. 

Let  ABO  he  a  triangle,  and  CD  a 
line  bisecting  the  angle  C.  Then 
we  are  to  prove  that 

CD'  +  (AD  x  DB)  =  ACx  OB. 
The  two  A's,  AOE  and  CDB,  are 
equiangular,  because  the  angles  E 
and  B  are  equal,  both  being  in  the 
same  segment,  and  the  [_  ACE  =  BCD,  by  hypothesis. 
Therefore,  (Th.  17,  Cor.  1,  B.  H), 

AC  :  CE  n  CD  :  CB. 
But  it  is  obvious  that  CE  =  CD  -f  DE,  and  by  substi- 
tuting this  value  of  CE,  in  the  proportion,  we  have, 
AC  :  CD  +  DE  ::  CD  :  CB. 
By  multiplying  extremes  and  means, 

UD2  +  (DE  x  CD)  =  ACx  CB. 
But  by  (Th.  17), 

DE  x  CD  =  AD  x  DB, 
and  substituting,  we  have, 

CD2  +  (AD  x  DB)  =  ACx  CB. 
Hence  the  theorem. 

THEOREM   XXI. 

The  rectangle  contained  by  the  two  diagonals  of  any  quad- 
rilateral inscribed  in  a  circle,  is  equivalent  to  the  sum  of  the 
tivo  rectangles  contained  by  the  opposite  sides  of  the  quadri- 
lateral. 

Let  ABCD  be  a  quadrilateral  inscribed  in  a  circle; 
then  we  are  to  prove  that 

AC  x  BD  =  (AB  x  DC)  -f  (AD  x  BC). 
From  Cy  draw  CE,  making  the  angle  DCE  equal  to 


BOOK    III.  107 

the  angle  A  OB;  and  as  the  angle  BAOis  equal  to  the 
angle  ODE,  both  being  in  the  same  seg- 
ment, therefore,  the  two  triangles,  DEC 
and  ABC,  are  equiangular,  and  we  have 
(Th.  17,  Cor.  1,  B.  II), 

AB  :  AC  ::  BE  :  DC    (1) 
The  two  A's,  ABC  and  BEC,  are 
equiangular;  for  the  \__DAC=  [__EBC, 
both  being  in  the  same  segment;  and  the  |_  BCA  = 
[_ECB,  for  BCE  =  BCA;  to  each  of  these  add  the  angle 
EC  A,  and  BCA  =  ECB;  therefore,   (Th.  17,   Cor.  1, 

B.  II), 

AB  :  AC  -.:  BE  :  BO    (2). 

By  multiplying  the  extremes  and  means  in  proportions 
(1)  and  (2),  and  adding  the  resulting  equations,  we  have, 

(AB  x  DC)  +  (AB  x  BO)  =  (BE  +  BE)  x  AC. 
But,  DE  +  BE  =  BB ;  therefore, 

(AB  x  DO)  +  (AD  x  BO)  =  AC  x  #Z). 

,   Cor.  When  two  adjacent  sides  of  the  quadrilateral  aio 
equal,  as  AB  and  BO,  then  the  resulting  equation  is, 

(AB  x  DC)  +  (AS  x  AD)  =  AC  x  BB; 
or,  AB  x  (BO  +  4i>)  =  AC  x  BB; 

or,  AB  :  AC  ::  BB  :  DC+  AB. 

That  is,  owe  of  the  two  equal  sides  of  the  quadrilateral 
is  to  the  adjoining  diagonal,  as  the  transverse  diagonal  is  to 
the  sum  of  the  two  unequal  sides. 

THEOREM   XXII. 

If  two  chords  intersect  each  other  at  right  angles  in  a  cir- 
cle, the  sum  of  the  squares  of  the  four  segments  thus  formed 
is  equivalent  to  the  square  of  the  diameter  of  the  circle. 

Let  AB  and   CD  be   two   chords,  intersecting   each 
other  at  right  angles.     Draw  BE  parallel  to  EB,  and 
draw  DF  and  A F.    Now,  we  are  to  prove  that 
~AE2  +~EB2  +~EC2  +  ED*  =TAF\ 


108 


GEOMETRY. 


As  BF  is  parallel  to  ED,  ABF  is  a 
riglit  angle,  and  therefore  AFis  a  diam- 
eter, (Th.  9).  Also,  because  BF  is 
parallel  to  CD,  CB  =  DF,  (Th.  13). 

Because  QEB  is  a  right  angle, 

OF2  +  FB2  =  OB2  =  DF2. 
Because  AFD  is  a  right  angle, 

~AF2+~ED2  =  AD2. 

Adding  these  two  equations,  we  have, 

OF2  +~FB2  +  AF2  +~ED2  =  DF2  +  ~AD2. 
But,  as  AF  is  a  diameter,  and  ADF  a  right  angle, 

(Th.9), ___ 

DT+AD2  =  AF2; 

therefore,     OF2  +  i£#2  +  AF2  +  ED2  -  27P2. 
Hence  the  theorem. 

Scholium.  —  If  two  chords  intersect  each  other  at  right  angles,  in  a 
circle,  and  their  opposite  extremities  be  joined,  the  two  chords  thus 
formed  may  make  two  sides  of  a  right-angled  triangle,  of  which  the 
diameter  of  the  circle  is  the  hypotenuse. 

For,  AD  is  one  of  these  chords,  and  CB  is  the  other ;  and  we  have 
shown  that  CB  =  DF;  and  AD  and  DF  are  two  sides  of  a  right- 
angled  triangle,  of  which  AF  is  the  hypotenuse ;  therefore,  AD  and 
CB  may  be  considered  the  two  sides  of  a  right-angled  triangle,  and 
AF  its  hypotenuse. 

THEOREM  XXIII. 

If  two  secants  intersect  each  other  at  right  angles,  the  sum 
of  their  squares,  increased  by  the  sum  of  the  squares  of  the 
two  segments  without  the  circle,  will  be  equivalent  to  the  square 
of  the  diameter  of  the  circle. 

Let  AF  and  ED  be  two  secants  in- 
tersecting at  right  angles  at  the  point 
E.  From  B,  draw  BF  parallel  to  CD, 
and  draw  AF  and  AD.  !Now  we  are  to 
prove  that 

FA2  +  ED2  +  EB2  +~E(f  =  AF2. 


BOOK  III.  109 

Because  BF  is  parallel  to  CB,  ABF  is  a  right  angle, 
and  consequently  A F  is  a  diameter,  and  BC  —  BF;  and 
because  AF  is  a  diameter,  ABF  is  a  right  angle.  As 
ABB  is  a  right  angle, 

~AE2+W52=AD2 

Also, FB2+JEC2=BC2=BF2 

By  addition,  A^2+WD2+^2+W2=A^2+BF2=AFi 
Hence  the  theorem. 


THEOREM  XXIV. 

If  perpendiculars  be  drawn  to  each  of  the  sides  of  a  plane 
triangle,  they  will,  when  sufficiently  produced,  meet  in  a  com- 
mon point. 

The  three  angular  points  of  a  triangle  are  not  in  the 
same  straight  line;  consequently  one  circumference, 
and  but  one,  may  be  made  to  pass  through  them. 

Conceive  a  triangle  to  be  thus  circumscribed.  The 
sides  of  the  triangle  then  become  chords  of  the  circum- 
scribing circle,  and  they  are  bisected  by  the  perpendicu- 
lar radii,  (Th.  6). 

Conversely:  The  perpendiculars  bisecting  the  three 
sides  of  a  triangle  will  meet  in  a  common  point,  and 
that  point  will  be  the  center  of  the  circumscribing  circle. 

Hence  the  theorem. 

THEOREM  XXV. 

The  sums  of  the  opposite  sides  of  a  quadrilateral  circum- 
scribing a  circle  are  equal. 

Let  ABCB  be  a  quadrilateral  circumscribed  about  a 
circle,  whose  center  is  0.     Then  we  are  to  prove  that 
AB  +  BC=AB  +  BC. 
From  the  center  of  the  circle  draw  OF  and  OF  to 
the  points  of  contact  of  the  sides  AB  and  BC.     Then, 
10 


110 


GEOMETRY. 


the  two  right-angled  triangles,  OEB  and  OFB,  are  equal, 
because  they  have  the  hypotenuse 
OB  common,  and  the  side  OF  = 
OE;  therefore,  BE  =  BF,  (Cor., 
Th.  23,  B.  I). 

In  like  manner  we  can  prove 
that 
AE=AH,  CF=  CG,  sindDG^DK 

Now,  taking  the  equation  BE  = 
BF,  and  adding  to  its  first  mem- 
ber CG,  and  to  its  second  the 
equal  line  OF.  we  have, 

BE  +  CG  =  BF  +  OF    (1) 

The  equation  AE=AE,  by  adding  to  its  first  member 
DG}  and  to  the  second  the  equal  line,  BE,  gives 
AE+BG=AE+BE    (2) 

By  the  addition  of  (1)  and  (2),  we  find  that 
BE  +  AE+  CG  +  BG  =  BF  +  CF+AH+BH. 

That  is,  AB  +  CD  +BC+  AD. 

Hence  the  theorem. 


BOOK   IV. 


Ill 


BOOK  IV. 


PROBLEMS. 

In  this  section,  we  have,  in  most  instances,  merely 
shown  the  construction  of  the  problem,  and  referred  to 
the  theorem  or  theorems  that  the  student  may  use,  to 
prove  that  the  object  is  attained  by  the  construction. 

In  obscure  and  difficult  problems,  however,  we  have 
gone  through  the  demonstration  as  though  it  were  a 
theorem. 

PROBLEM    I. 


To  bisect  a  given  finite  straight  line. 

Let  AB  be  the  given  line,  and  from 
its  extremities,  A  and  B,  with  any 
radius  greater  than  one  half  of  AB, 
(Postulate  3),  describe  arcs,  cutting  A  — 
each  other  in  n  and  m.  Draw  the  line 
nm ;  and  (7,  where  it  cuts  AB,  will  be 
the  middle  of  the  given  line. 

Proof,  (B.  I,  Th.  18,  Sch.  2). 

PROBLEM   II. 

To  bisect  a  given  angle. 

Let  ABO  be  the  given  angle.  With  any 
radius,  and  B  as  a  center,  describe  the  arc 
AC.  From  A  and  <7,  as  centers,  with  a 
radius  greater  than  one  half  of  AG,  de- 
scribe arcs,  intersecting  in  n ;  join  B  and  n ; 
the  joining  line  will  bisect  the  given  angle. 

Proof,  (Th.  21,  B.  I). 


•k 


x 


A 


OF  THE  '  *V 


lih!l\/rr>~ 


112 


GEOMETRY. 


A  n 


m  B 


Proof, 


PROBLEM    III. 

From  a  given  point  in  a  given  line,  to  draw  a  perpendicular 
to  that  line. 

Let  AB  be  the  given  line,  and 
O  the  given  point.  Take  n  and  m, 
equal  distances  on  opposite  sides 
of  0;  and  with  the  points  m  and 
n,  as  centers,  and  any  radius 
greater  than  nO  or  mO,  describe 
arcs  cutting  each  other  in  S,  Draw 
SO,  and  it  will  be  the  perpendicular  required. 
(B.  I,  Th.  18,  Sch.  2). 

The  following  is  another  method, 
which  is  preferable,  when  the  given 
point,  0,  is  at  or  near  the  end  of  the 
line. 

Take  any  point,  0,  which  is  mani- 
festly one  side  of  the  perpendicular, 
as  a  center,  and  with  00  as  a  radius,  describe  a  circum- 
ference, cutting  AB  in  m  and  0.  Draw  mn  through  the 
points  m  and  0,  and  meeting  the  arc  again  in  n ;  mn  is 
then  a  diameter  to  the  circle.  Draw  On,  and  it  will  be 
the  perpendicular  required.     Proof,  (Th.  9,  B.  III). 


A  m 


PROBLEM    IV. 

From  a  given  point  without  a  line,  to  draw  a  'perpendicular 
to  that  line. 

Let  AB  be  the  given  line,  and  O 
the  given  point.  From  0  draw  any 
oblique  line,  as  On,  Find  the  mid- 
dle point  of  On  by  Problem  1,  and 
with  that  point,  as  a  center,  describe 
a  semicircle,  having  On  as  a  diam- 
eter. From  m,  where  this  semi-cir- 
cumference cuts  AB,  draw  Om,  and  it  will  be  the  perpen- 
dicular required.     Proof,  (Th.  9,  B.  III). 


m     B 


BOOK  IV. 


113 


PROBLEM   V. 

At  a  given  point  in  a  line,  to  construct  an  angle  equal  to 
a  given  angle. 

Let  A  be  the  point  given  in  the  line 
AB,  and  DOE  the  given  angle. 

With  C  as  a  center,  and  any  radius, 
OF,  draw  the  arc  FD. 

With  A  as  a  center,  and  the  radius 
A F=  OF,  describe  an  indefinite  arc ;  and 
with  J7  as  a  center,  and  FG-  as  a  radius, 
equal  to  FD,  describe  an  arc,  cutting  the 
other  arc  in  Gr,  and  draw  A  G-;  GrAF  will  be  the  angle 
required.    Proof,  (Th.  5,  B.  III). 


PROBLEM   VI. 

From  a  given  point,  to  draw  a  line  parallel  to  a  given  line. 

Let  A  be  the  given  point,  and  BO  the 
given  line.  Draw  A  C,  making  an  angle, 
A  OB;  and  from  the  given  point,  A,  in 
the  line  AC,  draw  the  angle  CAD  = 
ACB,  by  Problem  5. 

Since  AJD  and  BO  make  the  same  angle  with  AC,  they 
are,  therefore,  parallel,  (B.  I,  Th.  7,  Cor.  1). 


PROBLEM   VII. 
To  divide  a  given  line  into  any  number  of  equal  parts. 

Let  AB  represent  the  given 
line,  and  let  it  be  required  to  di- 
vide it  into  any  number  of  equal 
parts,  say  live.  From  one  end  of 
the  line  A,  draw  AJD,  indefinite 
in  both  length  and  position.  Take 
any  convenient  distance  in  the  di- 
10*  h 


114 


GEOMETRY. 


viders,  as  Aa,  and  set  it  off  on  the  line  AD,  thus  making 
the  parts  Aa,  ab,  be,  etc.,  equal.  Through  the  last  point, 
e,  draw  EB,  and  through  the  points  a,  b,  c,  and  d,  draw 
parallels  to  eB,  by  Problem  6 ;  these  parallels  will  divide 
the  line  as  required.    Proof,  (Th.  17,  Book  IT). 


PROBLEM   VIII, 


To  find  a  third  proportional  to  two  given  lines. 


Let  AB  and  A  0  be  any  two  lines. 
Place  them  at  any  angle,  and  draw 
CB.  On  the  greater  line,  AB,  take 
AD  —  AO,  and  through  D,  draw 
DE  parallel  to  BO',  AE  is  the  third 
proportional  required. 

Proof,  (Th.  17,  B.  n). 


PROBLEM   IX. 

To  find  a  fourth  proportional  to  three  given  lines. 

Let  AB,  AC,  AD,  represent  the  A"~ 
three  given  lines.  Place  the  first 
two  at  any  angle,  as  BAO,  and  draw 
BO.  On  AB  place  AD,  and  from 
the  point  D,  draw  DE  parallel  to 
BO,  by  Problem  6 ;  AE  will  be  the 
fourth  proportional  required. 

Proof,  (Th.  17,  B.  II). 


PROBLEM   X. 

To  find  the  middle,  or  mean  proportional,  between  two  given 
lines. 


BOOK    IV. 


115 


Place  AB  and  BC  in  one  right 
line,  and  on  A  C,  as  a  diameter,  de- 
scribe a  semicircle,  (Postulate  3), 
and  from  the  point  B,  draw  BD  at 
right  angles  to  AC,  (Problem  3); 
BD  is  the  mean  proportional  re- 
quired. 

Proof,  (B.  m,  Th.  17,  Cor.). 

PROBLEM  XI. 

To  find  the  center  of  a  given  circle. 

Draw  any  two  chords  in  the  given  cir- 
cle, as  AB  and  CD,  and  from  the  middle 
points,  m  and  n,  draw  perpendiculars  to 
AB  and  CD ;  the  point  at  which  these 
two  perpendiculars  intersect  will  be  the 
center  of  the  circle. 

Proof,  (B.  m,  Th.  1,  Cor.). 


PROBLEM  XII. 

To  draw  a  tangent  to  a  given  circle,  from  a  given 
either  in  or  without  the  circumference  of  the  circle. 

When  the  given  point  is  in  the  cir- 
cumference, as  A,  draw  the  radius  A  C, 
and  from  the  point  A,  draw  AB  per- 
pendicular to  AC;  AB  is  the  tangent 
required. 

Proof,  (Th.  4,  B.  HI). 

"When  the  given  point  is  without 
the  circle,  as  A,  draw  AC  to  the 
center  of  the  circle ;  on  i(J,  as  a 
diameter,  describe  a  semicircle ;  and 
from  B,  where  the  semi-circumfer- 
ence cuts  the  given  circumference, 
draw  AB,  and  it  will  be  tangent  to  the  circle. 

Proof,  (Th.  9,  B.  Ill),  and,  (Th.  4,  B.  III). 


point, 


116 


GEOMETRY. 


PROBLEM    XIII. 

On  a  given  line,  to  describe  a  segment  of  a  circle,  that  shall 
contain  an  angle  equal  to  a  given  angle. 

Let  AB  be  the  given 
line,  and  O  the  given 
angle.  At  the  ends  of 
the  given  line,  form  angles 
DAB,  DBA,  each  equal 
to  the  given  angle,  O. 
Then  draw  AE  and  BE 
perpendiculars  to  AD  and  BD ;  and  with  E  as  a  center, 
and  EA,  or  EB,  as  a  radius,  describe  a  circle ;  then  AFB 
will  be  the  segment  required,  as  any  angle  F,  made  in 
it,  will  be  equal  to  the  given  angle,  O. 

Proof,  (Th.  11,  B.  HI),  and  (Th.  8,  B.  LEI). 

PROBLEM    XIV.  ■ 

From  any  given  circle  to  cut  a  segment,  that  shall  contain 
a  given  angle. 

Let  0  be  the  given  angle.  Take 
any  point,  as  A,  in  the  circumfer- 
ence, and  from  that  point  draw  the 
tangent  AB ;  and  from  the  point 
A,  in  the  line  AB,  construct  the 
angle  BAD  =  0,  (Problem  5),  and 
AED  is  the  segment  required. 

Proof,  (Th.  11,  B.  IH),  and  (Th.  8,  B.  III). 


PROBLEM   XV. 

To  construct  an  equilateral  triangle  on  a  given  straight  line. 

Let  AB  be  the  given  line;  from 
the  extremities  A  and  B,  as  centers, 
with  a  radius  equal  to  AB,  describe  arcs 
cutting  each  other  at  0.  From  0,  the 
point  of  intersection,  draw  OA  and  CB; 
ABO  will  be  the  triangle  required. 

The  construction  is  a  sufficient  demonstration.    Or,  (Ax.  1). 


BOOK    IV. 


117 


PROBLEM    XVI. 

To  construct  a  triangle,  having  its  three  sides  equal 
given  lines,  any  two  of  which  shall  be  greater  than  the 

Let  AB,  OB,  and  EF,  represent  the       E 

three  lines.   Take  any  one  of  them,  as  c 

AB,  to  be  one  side  of  the  triangle.  From 
A,  as  a  center,  with  a  radius  equal  to  CD, 
describe  an  arc ;  and  from  B,  as  a  center, 
with  a  radius  equal  to  EF,  describe  an- 
other arc,  cutting  the  former  in  n.  Draw 
An  and  Bn,  and  AnB  will  be  the  A  re- 
quired.     Proof,  (Ax.  1). 


to  three 
third. 

F 

D 


PROBLEM   XVII. 

To  describe  a  square  on  a  given  line. 

Let  AB  be  the  given  line ;  and  from  the 
extremities,  A  and  B,  draw  A  0  and  BB  per-     c 
pendicular  to  AB.     (Problem  3.) 

From  A,  as  a  center,  with  AB  as  radius, 
strike  an  arc  across  the  perpendicular  at  C;      t 
and  from  O  draw  OB  parallel  to  AB ;  AOBB 
is  the  square  required.      Proof,  (Th.  26,  B.  I). 


PROBLEM   XVIII. 

To  construct  a  rectangle,  or  a  parallelogram,  whose  adja- 
cent sides  are  equal  to  two  given  lines. 

Let  AB  and  A  0  be  the  two  given       A c 

lines.     From  the  extremities  of  one        A # 

line,  draw  perpendiculars  to  that  line,  as  in  the  last  prob- 
lem ;  and  from  these  perpendiculars,  cut  off  portions 
equal  to  the  other  line ;  and,  by  a  parallel,  complete  the 
figure. 


118  GEOMETRY. 

When  the  figure  is  to  be  a  parallelogram,  with  oblique 
angles,  describe  the  angles  by  Problem  5.  Proof,  (Th. 
26,  B.  I). 

PROBLEM   XIX. 

To  describe  a  rectangle  that  shall  be  equivalent  to  a  given 
square,  and  have  a  side  equal  to  a  given  line. 

Let  AB  be  a  side  of  the  given  square,        c D 

and  CD  one  side  of  the  required  rect-        A B 

angle.  E p 

Find  the  third  proportional,  FF,  to  CD  and  AB,  (Prob- 
lem 8).     Then  we  shall  have 

CD  :  AB  ::  AB  :  FF. 

Construct  a  rectangle  with  the  two  given  lines,  CD 
and  FF,  (Problem  18),  and  it  will  be  equal  to  the  given 
square,  (Th.  3,  B.  II). 

PROBLEM   XX. 

To  construct  a  square  that  shall  be  equivalent  to  the  differ- 
ence of  two  given  squares. 

Let  A  represent  a  side  of  the  greater  of  two  given 
squares,  and  B  a  side  of  the  less  square. 

On  A,  as  a  diameter,  describe  a 
semicircle,  and  from  one  extremity, 
p,  as  a  center,  with  a  radius  equal  to 
B,  describe  an  arc,  n,  and,  from  the 
point  where  it  cuts  the  circumference,  — - — 

draw  mn  and  np ;  np  is  the  side  of 
a  square,  which,  when  constructed, 
will  be  equal  to  the  difference  of  the  two  given  squares, 
(Problem  17).     Proof,  (Th.  9,  B.  Ill,  and  Th.  S6,  B.  I.) 

To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares,  we  have  only  to  draw  through  any  point 
two  lines  at  right  angles,  and  lay  off  on  one  a  distance 
equal  to  the  side  of  one  of  the  squares,  and  on  the  other 


BOOK    IV. 


119 


a  distance  equal  to  the  side  of  the  other.  The  straight 
line  connecting  the  extremities  of  these  lines  will  be  the 
side  of  the  required  square,  (Th.  36,  B.  I). 


PROBLEM  XXI. 

To  divide  a  given  line  into  two  parts,  which  shall  be  in  the 
ratio  of  two  other  given  lines. 


M^ 


Ni- 


Let  AB  be  the  line  A  ~HB 

to  be  divided,  and  M 
and  N  the  lines  hav- 
ing the  ratio  of  the 
required  parts  of  AB. 
From  the  extremity 
A  draw  AZ),  making 
any  angle  with  AB, 
and  take  AC  =  M, 
and  CD  =  N.  Join 
the  points  D  and  B 
by  a  straight  line, 
and  through  C  draw 
Ca  parallel  to  BD. 
Then  will  the  point  Gf  divide  the  line  AB  into  parts 
having  the  required  ratio.  (Proof,  Th.  17,  B.  II). 

Or,  having  drawn  AD,  lay  off  A C  =  M,  and  through 
B  draw  B  V  parallel  to  AD,  making  it  equal  to  N,  and 
join  C  and  V  by  a  line  cutting  AB  in  the  point  (7. 

Then  the  two  triangles  ACGr  and  GrBV  are  equiangu- 
lar and  similar,  and  their  homologous  sides  give  the 
proportion, 

Aa  :  GB  :  AC  ::  BV  ::  Mi  N 

The  line  AB  is  therefore  divided,  at  the  point  Q-,  into 
parts  which  are  in  the  ratio  of  the  lines  M  and  N", 


120 


GEOMETKY. 


PROBLEM  XXII. 

To  divide  a  given  line  into  any  number  of  parts,  having  to 
each  other  the  ratios  of  other  given  lines. 

Let  AB  be  the  given  M 
line  to  be  divided,  and  Nl 
M,  JST,  P,  etc.,  the  lines   p> 
to  which  the  parts  of 
AB  are  to  be  propor- 
tional. 

Through  the  point  A 
draw  an  indefinite  line,  making,  with  AB,  any  conve- 
nient angle,  and  on  this  line  lay  off  from  A  the  lines  M, 
JV,  P,  etc.,  successively.  Join  the  extremity  of  the  last 
line  to  the  point  B  by  a  straight  line,  parallel  to  which 
draw  other  lines  through  the  points  of  division  of  the 
indefinite  line,  and  they  will  divide  the  line  AB  at  the 
points  0,  D,  etc.,  into  the  required  parts.  (Proof,  Th.  17, 
B.  II). 

PROBLEM    XXIII. 

To  construct  a  square  that  shall  be  to  a  given  square,  as  a 
line,  M,  to  a  line,  N. 

Place  M  and  N  in  a  line,  and 
on  the  sum  describe  a  semicir- 
cle. From  the  point  where  the 
two  lines  meet,  draw  a  perpen- 
dicular to  meet  the  circumfer- 
ence in  A.  Draw  Am  and  An, 
and  produce  them  indefinitely.  On  Am  or  Am  produced, 
take  AB  =  to  the  side  of  the  given  square ;  and  from 
B,  draw  BO  parallel  to  mn;  A  0  is  a  side  of  the  required 
square. 

For,  Am  :A^2::  AB2  : ~AC\  (Th.  17,  B.  II). 

Also,  Am  :An  ::M     :  JV,      (Th. 25, B.II. Sch.). 

Therefore,  A&  :  A  C2 : :  M     :  1ST,      (Th.  6,  B.  II). 


BOOK    IV. 


121 


PROBLEM    XXIV. 

To  cut  a  line  into  extreme  and  mean  ratio  ;  that  is,  so  that 
the  whole  line  shall  be  to  the  greater  part,  as  that  greater  part 
is  to  the  less. 

Remark.  —  The  geometrical  solution  of  this  problem  is  not  imme- 
diately apparent,  but  it  is  at  once  suggested  by  the  form  of  the  equa- 
tion, which  a  simple  algebraic  analysis  of  its  conditions  leads  to. 

Bepresent  the  line  to  be  divided  by  2a,  the  greater 
part  by  x,  and  consequently  the  other,  or  less  part,  by 
2a  —  x. 

Now,  the  given  line  and  its  two  parts  are  required,  to 
satisfy  the  following  proportion : 

2a  :  x  : :  x  :  2a  —  x 

whence,  x2  =  4a2  —  2ax 

By  transposition,     x2  -f  2ax  =  4a2  =  (2a)2 

If  we  add  a2  to  both  members  of  this  equation,  we 
shall  have, 

x2  -f  2ax  +  a2  =  (2af  +  a2 

,or,  (x  +df  =  (2af  +  a2 

This  last  equation  indicates  that  the  lines  represented 
by  (x  +  a),  2a,  and  a,  are  the  three  sides  of  a  right- 
angled  triangle,  of  which  (x  +  a)  is  the  hypotenuse,  the 
given  line,  2a,  one  of  the  sides,  and  its  half,  a,  the  other. 

Therefore,  let  AB  represent  the 
given  line,  and  from  the  extremity,  B, 
draw  BO  at  right  angles  to  AB,  and 
make  it  equal  to  one  half  of  AB. 

With  0,  as  a  center,  and  radius  CB, 
describe  a  circle.  Draw  A  0  and  pro- 
duce it  to  F.  With  A  as  a  center 
and  AB  as  a  radius,  describe  the  arc 
BE)  this  arc  will  divide  the  line  AB, 
as  required. 

We  are  now  to  prove  that 

AB  :  AB  : :  AB  :  EB 
11 


122  GEOMETRY. 

By  Scholium  to  Th.  18,  B.  m,  we  have, 

AF  x  AD  =  AB? 
or,  AF  :  AB  : :  AB  :  AD 

Then,  (by  Cor.,  Th.  8,  Book  II),  we  may  have, 
(AF— AB)  :  AB  ::  (AB  —  AD)  :  AD 
Since         CB  =  \AB  =  JD.F;  therefore,  AB  =  DJ7. 
Hence,  AF  —  AB  =  AF  —  DF  =  AD  =  AF. 

Therefore,    AF  :  AB  ::  FB  :  AF 
By  taking  the  extremes  for  the  means,  we  have, 
AB  :  AF  : :  AF  :  FB. 

0 

PROBLEM   XXV. 

To  describe  an  isosceles  triangle,  having  its  two  equal  angles 
each  double  the  third  angle,  and  the  equal  sides  of  any  given 
length. 

Let  AB  be  one  of  the  equal  sides  of 
the  required  triangle;  and  from  the 
point  A,  with  the  radius  AB,  describe 
an  arc,  BD. 

Divide  the  line  AB  into  extreme  and 
mean  ratio  by  the  last  problem,  and  sup- 
pose C  the  point  of  division,  and  A  0  the 
greater  segment. 

From  the  point  B,  with  AC,  the  greater  segment,  as  a 
radius,  describe  another  arc,  cutting  the  arc  BD  in  D. 
Draw  BD,  DC,  and  DA.  The  triangle  ABD  is  the  tri- 
angle required. 

As  AC  —  BD,  by  construction ;  and  as  AB  is  to  A C 
as  A C is  to  B C,  by  the  division  of  AB;  therefore 
AB  :  BD  : :  BD  :  BC 

Now,  as  the  terms  of  this  proportion  are  the  sides  of 
the  two  triangles  about  the  common  angle,  B,  it  follows, 
(Cor.  2,  Th.  17,  B.  II),  that  the  two  triangles,  ABD  and 


BOOK   IV.  123 

BBC,  are  equiangular;  but  the  triangle  ABB  is  isos- 
celes; therefore,  BBC  is  isosceles  also,  and  BB  =  BO; 
but  BB  m  AC:  hence,  BC  =  AC,  (Ax.  1),  and  the  tri- 
angle ACB  is  isosceles,  and  the  [_  CBA  =  [_  J..  But 
the  exterior  angle,  BCB  m  CBA  +  A,  (Th.  12,  B.  I). 
Therefore,  [_BCB,  or  its  equal  \__B  =  L CBA  ■+■[__ A;  or 
the  angle  B  =  2[__A.  Hence,  the  triangle  ABB  has  each 
of  its  angles,  at  the  base,  double  of  the  third  angle. 

Scholium. — As  the  two  angles,  at  the  base  of  the  triangle  ABD,  are 
equal,  and  each  is  double  the  angle  A,  it  follows  that  the  sum  of  the 
three  angles  is  Jive  times  the  angle  A.  But,  as  the  three  angles  of  every 
triangle  are  always  equal  to  two  right  angles,  or  180°,  the  angle  A 
must  be  one  fifth  of  two  right  angles,  or  36° ;  therefore,  BD  is  a  chord 
of  36°,  when  AB  is  a  radius  to  the  circle ;  and  ten  such  chords  would 
extend  exactly  round  the  circle,  or  would  form  a  decagon. 


PROBLEM  XXVI. 

Within  a  given  circle  to  inscribe  a  triangle,  equiangular  to 
a  given  triangle. 

Let  ABC  be  the  circle,  and 
ale  the  given  triangle.  From 
any  point,  as  A,  draw  BB  tan- 
gent to  the  given  circle  at  A, 
(Problem  12). 

From  the  point  A,  in  the  line 
AB,  lay  off  the  angle  BAC  = 
the  angle  b,  (Problem  5),  and  the  angle  BAB  =  the  angle 
c,  and  draw  BC 

The  triangle  ABC  is  inscribed  in  the  circle;  it  is  equi- 
angular to  the  triangle  abc,  and  hence  it  is  the  triangle 
required. 

Proof,  (Th.  12,  B.  III). 


124 


GEOMETRY. 


PROBLEM   XXVII. 

To  describe  a  regular  pentagon  in  a  given  circle. 

1st.  Describe  an  isosceles  tri- 
angle, abc,  having  each  of  the 
equal  angles,  b  and  c,  double  the 
third  angle,  a,  by  Problem  25. 

2d.    Inscribe    the     triangle, 
ABO,  in  the  given  circle,  equi- 
angular to  the  triangle  abc,  by 
Problem  26  ;  then  each  of  the  angles,  B  and  0,  is  double 
the  angle  A. 

3d.  Bisect  the  angles  B  and  0,  by  the  lines  BB  and 
OE,  (Problem  2),  and  draw  AE,  EB,  CB,  BA;  and  the 
figure  AEBCB  is  the  pentagon  required. 

By  construction,  the  angles  BAG,  ABB,  BBC,  BOE, 
EGA,  are  all  equal ;  therefore,  (B.  HI,  Th.  9,  Scho.),  the 
arcs,  BO,  AB,  BO,  AE,  and  EB,  are  all  equal;  and  if 
the  arcs  are  equal,  the  chords  AE,  EB,  etc.,  are  equal. 

Scholium. — The  arc  subtended  by  one  of  the  sides  of  a  regular  pen- 

360° 
tagon,  being  one  fifth  of  the  whole  circumference,  is  equal  to  — — =72°* 

PROBLEM   XXVIII. 

To  describe  a  regular  hexagon  in  a  circle. 

Draw  any  diameter  of  the  circle,  as 
AB,  and  from  one  extremity,  B,  draw 
BB  equal  to  BO,  the  radius  of  the 
circle.  The  arc,  BB,  will  be  one  sixth 
part  of  the  whole  circumference,  and 
the  chord  BB  will  be  a  side  of  the  regu- 
lar polygon  of  six  sides. 

In  the  A  OBB,  as  OB  =  OB,  and  BB  =  OB  by  con- 
struction, the  A  is  equilateral,  and  of  course  equiangular. 

Since  the  sum  of  the  three  angles  of  every  A  is  equal 
to   two   right    angles,    or    to    180    degrees,   when   the 


E^- 

— ^p 

/  /       \ 

BOOK   IV. 


125 


three  angles  are  equal  to  one  another,  each  one  of  them 
must  be  60  degrees ;  but  60  degrees  is  a  sixth  part  of 
360  degrees,  the  whole  number  of  degrees  in  a  circle ; 
therefore,  the  arc  whose  chord  is  equal  to  the  radius,  is  a 
sixth  part  of  the  circumference ;  and,  if  a  polygon  of  six 
equal  sides  be  inscribed  in  a  circle,  each  side  will  be 
equal  to  the  radius. 

Scholium.  —  Hence,  as  BD  is  the  chord  of  60°,  and  equal  to  BC  qt 
CD,  we  say  generally,  that  the  chord  of  60°  is  equal  to  radius. 

PROBLEM  XXIX. 

To  find  the  side  of  a  regular  polygon  of  fifteen  sides,  which 
may  be  inscribed  in  any  given  circle. 

Let  CB  be  the  radius  of  the  given 
circle;  divide  it  into  extreme  and 
mean  ratio,  (Problem  24),  and  make 
BD  equal  to  CB,  the  greater  part; 
then  BD  will  be  a  side  of  a  regular 
polygon  of  ten  sides,  (Scholium  to 
Problem  25).  Draw  BA  =  to  CB,  and 
it  will  be  a  side  of  a  polygon  of  six  sides.  Draw  DA, 
and  that  line  must  be  the  side  of  a  polygon  which  cor- 
responds to  the  arc  of  the  circle  expressed  by  \  less  ^, 
of  the  whole  circumference ;  or  J  —  -^  =  g%  =  T^ ;  that 
is,  one-fifteenth  of  the  whole  circumference ;  or,  DA  is 
a  side  of  a  regular  polygon  of  15  sides.  But  the  15th 
part  of  360°  is  24°  ;  hence  the  side  of  a  regular  inscribed 
polygon  of  fifteen  sides  is  the  chord  of  an  arc  of  24°. 


PROBLEM    XXX 


In  a  given  circle  to  inscribe  a  regular  polygon  of  any  num- 
ber of  sides,  and  then  to  circumscribe  the  circle  by  a  similar 
polygon. 

11* 


126 


GEOMETRY. 


Let  the  circumference  of  the  circle,  whose  center  is  0, 
be  divided  into  any  number  of  equal  arcs,  as  AmB,  Bw(7, 
OoD,  etc. ;  then  will  the  polygon  abode,  etc.,  bounded  by 
the  chords  of  these  arcs,  be  regu- 
lar and  inscribed ;  and  the  poly- 
gon ABODE,  etc.,  bounded  by 
the  tangents  to  these  arcs  at  their 
middle  points  m,  n,  o,  etc,  be  a 
similar  circumscribed  polygon. 

First  —  The  polygon  abode, 
etc.,  is  equilateral,  because  its 
sides   are  the  chords   of  equal 

arcs  of  the  same  circle,  (Th.  5,  B.  Ill) ;  and  it  is  equi- 
angular, because  its  angles  are  inscribed  in  equal  segments 
of  the  same  circle,  (Th.  8,  B.  III).  Therefore  the  poly- 
gon is  regular,  (Def.  14,  B.  Ill),  and  it  is  inscribed,  since 
the  vertices  of  all  its  angles  are  in  the  circumference  of 
the  circle,  (Def.  13,  B.  HI). 

Second. — If  we  draw  the  radius  to  the  point  of  tangency 
of  the  side  AB  of  the  circumscribed  polygon,  this  radius 
is  perpendicular  to  AB,  (Th.  4,  B.  Ill),  and  also  to  the 
chord  ah,  (B.  Ill,  Th.  1,  Cor.) ;  hence  AB  is  parallel  to  ah, 
and  for  the  same  reason  BO  is  parallel  to  bo ;  therefore 
the  angle  ABO  is  equal  to  the  angle  abo,  (Th.  8,  B.  I). 
In  like  manner  we  may  prove  the  other  angles  of  the 
circumscribed  polygon,  each  equal  to  the  corresponding 
angle  of  the  inscribed  polygon.  These  polygons  are 
therefore  mutually  equiangular. 

Again,  if  we  draw  the  radii  Om  and  On,  and  the  line  OB, 
the  two  A's  thus  formed  are  right-angled,  the  one  at  m 
and  the  other  at  n,  the  side  OB  is  common  and  Om  is 
equal  to  On ;  hence  the  difference  of  the  squares  described 
on  OB  and  Om  is  equivalent  to  the  difference  of  the 
squares  described  on  OB  and  On.  But  the  first  difference 
is  equivalent  to  the  square  described  on  Bm,  and  the 
second  difference  is  equivalent  to  the  square  described 


BOOK    IV.  127 

on  Bn ;  hence  Bm  is  equal  to  Bn,  and  the  two  right- 
angled  triangles  are  equal,  (Th.  20,  B.  I),  the  angle  BOm 
opposite  the  side  Bm  being  equal  to  the  angle  BOn,  op- 
posite the  equal  side  Bn.  The  line  OB  therefore  passes 
through  the  middle  point  of  the  arc  mbn ;  but  because  m 
and  n  are  the  middle  points  of  the  equal  arcs  amb  and 
bne,  the  vertex  of  the  angle  abe  is  also  at  the  middle 
point  of  the  arc  mbn.  Hence  the  line  OB,  drawn  from 
the  center  of  the  circle  to  the  vertex  of  the  angle  ABO, 
also  passes  through  the  vertex  of  the  angle  abc.  By  pre- 
cisely the  same  process  of  reasoning,  we  may  prove  that 
00  passes  through  the  point  c,  OD  through  the  point  d, 
etc. ;  hence  the  lines  joining  the  center  with  the  vertices 
of  the  angles  of  the  circumscribed  polygon,  pass  through 
the  vertices  of  the  corresponding  angles  of  the  inscribed 
polygon ;  and  conversely,  the  radii  drawn  to  the  vertices 
of  the  angles  of  the  inscribed  polygon,  when  produced, 
pass  through  the  vertices  of  the  corresponding  angles 
of  the  circumscribed  polygon. 

Now,  since  ab  is  parallel  to  AB,  the  similar  A's  abO 
and  ABO,  give  the  proportion 

Ob  :  OB  ::  ab  :  AB, 

and  the  A's, bcO  and  BOO,  give  the  proportion 

Ob  :   OB  : :  be  :  BO. 

As  these  two  proportions  have  an  antecedent  and  con- 
sequent, the  same  in  both,  we  have,  (Th.  6,  B.  II), 

ab  :  AB  : :  be  :  BO. 

In  like  manner  we  may  prove  that 

be  :  BO  : :  cd  :  OB,  etc.,  eta 

The  two  polygons  are  therefore  not  omy  equiangular, 
but  the  sides  about  the  equal  angles,  taken  in  the  same 
order,  are  proportional ;  they  are  therefore  similar,  (Def. 
16,  B.  n). 


128  GEOMETRY. 

Cor.  1.  To  inscribe  any  regular  polygon  in  a  circle,  we 
have  only  to  divide  the  circumference  into  as  many  equal 
parts  as  the  polygon  is  to  have  sides,  and  to  draw  the 
chords  of  the  arcs ;  hence,  in  a  given  circle,  it  is  possible 
to  inscribe  regular  polygons  of  any  number  of  sides 
whatever.  Having  constructed  any  such  polygon  in  a 
given  circle,  it  is  evident,  that  by  changing  the  radius  of 
the  circle  without  changing  the  number  of  sides  of  the 
polygon,  it  may  be  made  to  represent  any  regular  poly- 
gon of  the  same  name,  and  it  will  still  be  inscribed  in  a 
circle.  As  this  reasoning  is  applicable  to  regular  poly- 
gons of  whatever  number  of  sides,  it  follows,  that  any 
regular  polygon  may  be  circumscribed  by  the  circumference 
of  a  circle. 

Cor.  2.  Since  ab,  be,  cd,  etc.,  are  equal  chords  of  the 
same  circle,  they  are  at  the  same  distance  from  the 
center,  (Th.  3,  B.  Ill) ;  hence,  if  with  0  as  a  center,  and 
Ot,  the  distance  of  one  of  these  chords  from  that  point, 
as  a  radius,  a  circumference  be  described,  it  will  touch 
all  of  these  chords  at  their  middle  points.  It  follows, 
therefore,  that  a  circle  may  be  inscribed  within  any  regular 
polygon. 

Scholium. — The  center,  0,  of  the  circle,  may  be  taken  as  the  center 
of  both  the  inscribed  and  circumscribed  polygons;  and  the  angle 
A  OB,  included  between  lines  drawn  from  the  center  to  the  extremities 
of  one  of  the  sides  AB,  is  called  the  angle  at  the  center.  The  perpen- 
dicular drawn  from  the  center  to  one  of  the  sides  is  called  the  Apothem 
of  the  polygon. 

Cor.  3.    The  angle  at  the  center  of  any  regular  polygon 

is  equal  to  four  right  angles  divided  by  the  number  of 

sides  of  the  polygon.     Thus,  if  n  be  the  number  of  sides 

of  the  polygon,  the  angle  at  the  center  will  be  expressed 

.     360° 

by . 

n 

Cor. 4.     If  the   arcs   subtended  by  the  sides  of  any 

regular  inscribed  polygon  be  bisected,  and  the  chords 

of  these  semi-arcs  be  drawn,  we  shall  have  a  regular 


BOOK   IV.  129 

inscribed  polygon  of  double  the  number  of  sides.  Thus, 
from  the  square  we  may  pass  successively  to  regular 
inscribed  polygons  of  8,  16,  32,  etc.,  sides.  To  get  the 
corresponding  circumscribed  polygons,  we  have  merely 
to  draw  tangents  at  the  middle  points  of  the  arcs  sub- 
tended by  the  sides  of  the  inscribed  polygons. 

Cor.  5.  It  is  plain  that  each  inscribed  polygon  is  but 
a  part  of  one  having  twice  the  number  of  sides,  while 
each  circumscribed  polygon  is  but  a  part  of  one  having 
one  half  the  number  of  sides. 


130 


GEOMETRY. 


BOOK  V 


ON  THE  PROPORTIONALITIES  AND  MEASUREMENT 
OF  POLYGONS  AND  CIRCLES. 


PROPOSITION  I.  — THEOREM. 

The  area  of  any  circle  is  equal  to  the  product  of  its  radius 
by  one  half  of  its  circumference. 

Let  OA  be  the  radius  of  a  circle, 
and  AB  a  very  small  portion  of  its 
circumference;  then  A  OB  will  be  a 
sector.  "We  may  conceive  the  whole 
circle  made  up  of  a  great  number  of 
such  sectors;  and  when  each  sector 
is  very  small,  the  arcs  AB,  BD,  etc., 
each  one  taken  separately,  may  be  considered  a  right 
line ;  and  the  sectors  CAB,  CBD,  etc.,  will  be  triangles. 
The  triangle,  AOB,  is  measured  by  the  product  of  the 
base,  AC,  multiplied  into  one  half  the  altitude,  AB,  (Th. 
33,  Book  I) ;  and  the  triangle  BOD  is  measured  by  the  pro- 
duct of  BO,  or  its  equal,  AO,  into  one  half  BD;  then  the 
area,  or  measure  of  the  two  triangles,  or  sectors,  is  the 
product  of  AO,  multiplied  by  one  half  of  AB  plus  one 
half  of  BD,  and  so  on  for  all  the  sectors  that  compose 
the  circle ;  therefore,  the  area  of  the  circle  is  measured 
by  the  product  of  the  radius  into  one  half  the  circumference. 


BOOK    V.  131 

PROPOSITION  II.  —  THEOREM. 

Circumferences  of  circles  are  to  one  another  as  their  radii, 
and  their  areas  are  to  one  another  as  the  squares  of  their 
radii. 

Let  CA  be  the  radius  of  a  circle, 
and  Oa  the  radius  of  another  circle. 
Conceive  the  two  circles  to  be  so 
placed  upon  each  other  so  as  to  have 
a  common  center. 

Let  AB  be  such  a  certain  definite 
portion  of  the  circumference  of  the 
larger  circle,  that  m  times  AB  will  represent  that  cir- 
cumference. 

But  whatever  part  AB  is  of  the  greater  circumference, 
the  same  part  ah'  is  of  the  smaller;  for  the  two  circles 
have  the  same  number  of  degrees,  and  are  of  course  sus- 
ceptible of  division  into  the  same  number  of  sectors. 
But  by  proportional  triangles  we  have, 
CA  :  Ca  : :  AB  :  ab 

Multiply  the  last  couplet  by  m,  (Th.  4,  B.  II),  and  we 
have 

CA  :  Ca  ::  mAB  :  mab. 

That  is,  the  radius  of  one  circle  is  to  the  radius  of  another, 
as  the  circumference  of  the  one  is  to  the  circumference  of  the 
other. 

To  prove  the  second  part  of  the  theorem,  let  C  repre- 
sent the  area  of  the  larger  circle,  and  c  that  of  the 
smaller ;  now,  whatever  part  the  sector  CAB  is  of  the 
circle  C,  the  sector  Cab  is  the  corresponding  part  of  the 
circle  c. 

That  is,  Cic       : :  CAB  :  Cab, 

but,  CAB  :  Cab  : :  (CAf  :  (Ca)2,    (Th.  20,  B.  II). 

Therefore,        C :  c       : :  (CAf  :  (Ca)2,    (Th.  6,  B.  H). 

That  is,  the  area  of  one  circle  is  to  the  area  of  another,  a% 


132  GEOMETEY. 

the  square  of  the  radius  of  the  one  is  to  the  square  of  the 
radius  of  the  other. 
Hence  the  theorem. 

Cor.     If  0  :  e  ::  (OAf    :  (Co)2, 

then,  0  :  c  ::  4  (Oaf  :  4  (Ob)2. 

But  4  (OAf  is  the  square  of  the  diameter  of  the  larger 
circle,  and  4  (Oaf  is  the  square  of  the  diameter  of  the 
smaller.  Denoting  these  diameters  respectively  by  D 
and  d,  we  have, 

O  :  c  : :  D2  :  d\ 

That  is,  the  areas  of  any  two  circles  are  to  each  other,  as 
the  squares  of  their  diameters. 

Scholium.  —  As  the  circumference  of  every  circle,  great  or  small,  is 
assumed  to  be  the  measure  of  360  degrees,  if  we  conceive  the  circum- 
ference to  be  divided  into  360  equal  parts,  and  one  such  part  repre- 
sented bjAB  on  one  circle,  or  ab  on  the  other,  AB  and  ab  will  be  very- 
near  straight  lines,  and  the  length  of  such  a  line  as  AB  will  be  greater 
or  less,  according  to  the  radius  of  the  circle ;  but  its  absolute  length 
cannot  be  determined  until  we  know  the  absolute  relation  between  the 
diameter  of  a  circle  and  its  circumference. 

PROPOSITION    III.  — THEOREM. 

When  the  radius  of  a  circle  is  unity,  its  area  and  semi- 
circumference  are  numerically  equal. 

Let  R  represent  the  radius  of  any  circle,  and  the  Greek 
letter,  *,  the  half  circumference  of  a  circle  whose  radius 
is  unity.  Since  circumferences  are  to  each  other  as  their 
radii,  when  the  radius  is  R,  the  semi-circumference  will 
be  expressed  by  «R. 

Let  m  denote  the  area  of  the  circle  of  which  R  is  the 
radius ;  then,  by  Theorem  1,  we  shall  have,  for  the  area 
of  this  circle,  *R?  =  m,  which,  when  R  =  1,  reduces  to 
ic  =  m. 

This  equation  is  to  be  interpreted  as  meaning  that  the 
semi-circumference  contains  its  unit,  the  radius,  as  many 


BOOK   V. 


133 


times  as  the  area  of  the  circle  contains  its  unit,  the 
square  of  the  radius. 

Remark.  —  The  celebrated  problem  of  squaring  the  circle  has  for  its 
object  to  find  a  line,  the  square  on  which  will  be  equivalent  to  the  area 
of  a  circle  of  a  given  diameter ;  or,  in  other  words,  it  proposes  to  find 
the  ratio  between  the  area  of  a  circle  and  the  square  of  its  radius. 

An  approximate  solution  only  of  this  problem  has  been  as  yet  dis- 
covered, but  the  approximation  is  so  close  that  the  exact  solution  is 
no  longer  a  question  of  any  practical  importance. 


PROPOSITION   IV.  — PROBLEM. 

Given,  the  radius  of  a  circle  unity,  to  find  the  areas  of 
regular  inscribed  and  circumscribed  hexagons. 

Conceive  a  circle  described  with  the  radius  QA,  and  in 
this  circle  inscribe  a  regular  polygon  of  six  sides  (Prob. 
28,  B.  IV),  and  each  side  will  be 
equal  to  the  radius  QA ;  hence, 
the  whole  perimeter  of  this  poly- 
gon must  be  six  times  the  ra- 
dius of  the  circle,  or  three  times 
the  diameter.  The  chord  bd  is 
bisected  by  QA.  Produce  Ob  and  Qd,  and  through  the 
point  Ay  draw  BD  parallel  to  bd ;  BD  will  then  be  a  side 
of  a  regular  polygon  of  six  sides,  circumscribed  about 
the  circle,  and  we  can  compute  the  length  of  this  line, 
BD,  as  follows :  The  two  triangles,  Cbd  and  QBD,  are 
equiangular,  by  construction ;  therefore, 

Oa  :  bd  ::  QA  :  BD. 

Now,  let  us  assume  QA  =  QD  =  the  radius  of  the 
circle,  equal  unity ;  then  bd  =  1,  and  the  preceding  pro- 
portion becomes     - 

Qa  :  1  ::  1  :  BD        (1) 
In  the  right-angled  triangle  Qad,  we  have, 

Qa2  +  ad2  =  Qd2,        (Th.  39,  B.  I). 
That  is,  Qa2  -f  J  =  1,  because  Qd  =  1,  and  ad  =  J. 

12 


134  GEOMETRY. 

Whence,  Ca  =  J  </3.  This  value  of  Ca,  substituted  in 
proportion  (1),  gives 

tf/S  :  1  : :  1  :  BD;  hence,  BD  =  JL 

But  the  area  of  the  triangle  Cbd  is  equal  to  bd  (=  1,) 
multiplied  by  \Ca  =  \  ^3 ;  and  the  area  of  the  triangle 
CBD  is  equal  to  BD  multiplied  by  |(M. 

Whence,  area,  Cbd  =  J  ^3, 

and,  area,  CJ2D  =    ,-=. 

V  3 

But  the  area  of  the  inscribed  polygon  is  six  times  that 
of  the  triangle  Cbd,  and  the  area  of  the  circumscribed 
polygon  is  six  times  that  of  the  triangle  CBD. 

Let  the  area  of  the  inscribed  polygon  be  represented 
by  p,  and  that  of  the  circumscribed  polygon  by  P. 

Thenp  =  |  V3,  andP  =  — = 
2  \/S 

Whence  p  :  P  : :  gv^  :  2^3  ::^:2::3:4::9  :  12 

p=^V3  =  2.59807621.    I 

Now,  it  is  obvious  that  the  area  of  the  circle  must  be 
included  between  the  areas  of  these  two  polygons,  and 
not  far  from,  but  somewhat  greater  than,  their  half  sum, 
which  is  3.03  -f  ;  and  this  may  be  regarded  as  the  first 
approximate  value  of  the  area  of  the  circle  to  the  radius 
unity. 

PROPOSITION   V.  — PROBLEM. 

Given,  the  areas  of  two  regular  polygons  of  the  same  num- 
ber of  sides,  the  one  inscribed  in  and  the  other  circumscribed 
about,  the  same  circle,  to  find  the  areas  of  regular  inscribed  and 
circumscribed  polygons  of  double  the  number  of  sides. 

Letp  represent  the  area  of  the  given  inscribed  polygon, 
and  P  that  of  the  circumscribed  polygon  of  the  same 


> 

2  x  3  =  2^3. 
V3 

3 
2  : 

:2::  3:4::  9  : 

2^3  =  3.46410161 

BOOK   V. 


135 


number  of  sides.  Also  denote  by  pf  the  area  of  the 
inscribed  polygon  of  double  the  number  of  sides,  and  by 
Pf  that  of  the  corresponding  circumscribed  polygon. 
Now,  if  the  arc  KAL  be  some  exact  part,  as  one-fourth, 
one  fifth,  etc.,  of  the  circumference  of  the  circle,  of  which 
Q  is  the  center  and  QA  the  radius,  then  will  KL  be  the 
side  of  a  regular  inscribed  polygon,  and  the  triangle 
KQL  will  be  the  same  part  of  the  whole  polygon  that 
the  arc  KAL  is  of  the  whole  circumference,  and  the 
triangle  QBB  will  be  a  like  part  of  the  circumscribed 
polygon.  Draw  QA  to  the  point  of  tangency,  and  bisect 
the  angles  ACB  and  AQB,  by  the  lines  QGr  and  QB,  and 
draw  KA. 

It  is  plain  that  the  triangle 
AOK  is  an  exact  part  of  the 
inscribed  polygon  of  double  the 
number  of  sides,  and  that  the 
A  EQG-  is  a  like  part  of  the  cir- 
cumscribed polygon  of  double 
the  number  of  sides.  Repre- 
sent the  area  of  the  A  LQK  by 
a,  and  the  area  of  the  A  BOB 
by  b,  that  of  the  A  ACK  by  x, 

and  that  of  the  A  BQGr  by  y,  and  suppose  the  A's,  KCL 
and  BBC,  to  be  each  the  nth  part  of  their  respective 
polygons. 

Then,  na  =  p,     nb  =  P,     2nx  =  pf, 

and,  Zny  =  Pf ; 

But,  by  (Th.  33,  B.  I),  we  have 

CM.MK=*a  (1) 
QA  .  AB  =  b  (2) 
QA  .  MK=2x      (3) 

Multiplying  equations  ( 1 )  and  ( 2 )?  member  by  member, 
we  have 

(CM  .  AB)  x  (QA  .  MK)  =ab        (4) 


136  GEOMETRY. 

From  the  similar  A's  CMK  and  CAD,  we  have 

CM  :  MK  : :  CA  :  AD 
whence  CM  .  AD  =  (Li  .  ME" 

But  from  equation  ( 3 )  we  see  that  each  member  of 
this  last  equation  is  equal  to  2x;  hence  equation  (4) 

becomes 

2#  .  2#  =  ab 

If  we  multiply  both  members  of  this  by  n*  =  n  .  n, 

we  shall  have 

4n2x*  =  na.nb  =  p.P 

or,  taking  the  square  root  of  both  members, 

2nx  =  s/pl? 

That  is,  the  area  of  the  inscribed  polygon  of  double  the 
number  of  sides  is  a  mean  proportional  between  the  areas  of 
the  given  inscribed  and  circumscribed  polygons  p  and  P. 

Again,  since  CE  bisects  the  angle  ACD,  we  have,  by, 
(Th.  24,  B.  II), 


AE  :  ED 


hence,    AE  :  AE  +ED 


CA  :  CD 
CM:  CK 
CM:  CA 

CM:  CM+  CA. 


Multiplying  the  first  couplet  of  this  proportion  by  CA, 
and  the  second  by  MK,  observing  that  AE  -f  ED  =  AD, 
we  shall  have 

AE.CA  :  AD.CA  ::  CM.MK  :  (CM  +  CA)  MK. 

But  AE.  CA  measures  the  area  of  the  A  CEG-,  which 
we  have  called  y,  AD.CA  =  A  CBD  =  5,  CM.MK  = 
A  CKL  =  a,  and  (CM  +  CA)MK  =  A  CMK,  and 
CAK  =  a  4-  2a:,  as  is  seen  from  equations  ( 1 )  and  ( 3 ). 
Therefore  the  above  proportion  becomes 
y  :  b  : :  a  :  a  +  2x. 

Multiplying  the  first  couplet  by  2n,  and  the  second  by 
n,  we  shall  have 


BOOK  V. 

That  is, 
whence, 

Zny  :  2nb  : :  na  :  na  +  2nx 
P'  :  2P    : :  p     :  p  +  p' 

137 


and  as  the  value  of  pf  has  been  previously  found  equal  to 
v'Pp,  the  value  of  Pf  is  known  from  this  last  equation, 
and  the  problem  is  completely  solved. 

PROPOSITION   VI.  — PROBLEM. 

To  determine  the  approximate  numerical  value  of  the  area 
of  a  circle,  when  the  radius  is  unity. 

We  have  now  found,  (Prob.  4),  the  areas  of  regular 
inscribed  and  circumscribed  hexagons,  when  the  radius 
of  the  circle  is  taken  as  the  unit ;  and  Prob.  5  gives  us 
formulae  for  computing  from  these  the  areas  of  regular 
inscribed  and  circumscribed  polygons  of  twelve  sides, 
and  from  these  we  may  again  pass  to  polygons  of 
twenty-four  sides,  and  so  on,  without  limit.  Now,  it  is 
evident  that,  as  the  number  of  sides  of  the  inscribed 
polygon  is  increased,  the  polygon  itself  will  increase, 
gradually  approaching  the  circle,  which  it  can  never  sur- 
pass. And  it  is  equally  evident  that,  as  the  number  of 
sides  of  the  circumscribed  polygon  is  increased,  the  poly- 
gon itself  will  decrease,  gradually  approaching  the  circle, 
less  than  which  it  can  never  become. 

The  circle  being  included  between  any  two  corres- 
ponding inscribed  and  circumscribed  polygons,  it  will 
differ  from  either  less  than  they  differ  from  each  other ; 
and  the  area  of  either  polygon  may  then  be  taken  as  the 
area  of  the  circle,  from  which  it  will  differ  by  an  amount 
less  than  the  difference  between  the  polygons. 

It  is  also  plain  that,  as  the  areas  of  the  polygons  ap- 
proach equality,  their  perimeters  will  approach  coinci- 
dence with  each  other,  and  with  the  circumference  of 
the  circle. 
12* 


138  GEOMETKY. 

Assuming  the  areas  already  found  for  the  inscribed 
and  circumscribed  hexagons,  and  applying  the  formulae 
of  Prob.  5  to  them  and  to  the  successive  results  ob- 
tained, we  may  construct  the  following  table : 

NUMBER  OF  SIDES.  INSCRIBED  POLYGONS.  CIRCUMSCRIBED  POLYGONS. 

6  ^v^  2.59807621  2^3=3.46410161 

19 

12  3  =  3.0000000        7r=—^=  3.2153904 

2+v/3 
6 
24      v^l+^f  =  3-1058286  3.1596602 

48  3.132628T  3.1460863 

96  3.1393554  3.1427106 

192  3.1410328  3.1418712 

384  3.1414519  3.1416616 

768  3.1415568  3.1416092 

1536  3.1415829  3.1415963 

3072  3.1415895  3.1415929 

6144  3.1415912  3.1415927 

Thus  we  have  found,  that  when  the  radius  of  a  circle  is 
1,  the  semi-circumference  must  be  more  than  3.1415912, 
and  less  than  3.1415927 ;  and  this  is  as  accurate  as  can 
be  determined  with  the  small  number  of  decimals  here 
used.  To  be  more  accurate  we  must  have  more  decimal 
places,  and  go  through  a  very  tedious  mechanical  opera- 
tion; but  this  is  not  necessary,  for  the  result  is  well 
known,  and  is  3.1415926535897,  plus  other  decimal  places 
to  the  100th,  without  termination.  This  result  was  dis- 
covered through  the  aid  of  an  infinite  series  in  the  Dif- 
ferential and  Integral  Calculus. 

The  number,  3.1416,  is  the  one  generally  used  in  prac- 
tice, as  it  is  much  more  convenient  than  a  greater  num- 
ber of  decimals,  and  it  is  sufficiently  accurate  for  all 
ordinary  purposes. 

In  analytical  expressions  it  has  become  a  general  cus- 
tom with  mathematicians  to  represent  this  number  by 


BOOK  V.  139 

the  Greek  letter  *,  and,  therefore,  when  any  diameter  of 
a  circle  is  represented  by  D,  the  circumference  of  the 
same  circle  must  be  *D.  If  the  radius  of  a  circle  is  re- 
presented by  M,  the  circumference  must  be  represented 
by  2*B. 

Scholium.  —  The  side  of  a  regular  inscribed  hexagon  subtends  an 
arc  of  60°,  and  the  side  of  a  regular  polygon  of  twelve  sides  subtends 
an  arc  of  30° ;  and  so  on,  the  length  of  the  arc  subtended  by  the  sides 
of  the  polygons,  varying  inversely  with  the  number  of  sides. 

Angles  are  measured  by  the  arcs  of  circles  included  between  their 
sides ;  they  may  also  be  measured  by  the  chords  of  these  arcs,  or  rather 
by  the  half  chords  called  sines  in  Trigonometry.  For  this  purpose,  it 
becomes  necessary  to  know  the  length  of  the  chord  of  every  possible 
arc  of  a  circle. 

PROPOSITION  VII.  — PROBLEM. 

Given,  the  chord  of  any  arc,  to  find  the  chord  of  one  half 
that  arc,  the  radius  of  the  circle  being  unity. 

Let  FE  be  the  given  chord,  and  draw 
the  radii  OA  and  OE,  the  first  perpen- 
dicular to  FE,  and  the  second  to  its  ex- 
tremity, E. 

Denote  FE  by  2c,  and  the  chord  of 
the  half  arc  AE  by  x. 

Then,   in  the  right-angled   triangle, 
DOE,   we  have  ~DC2  =  OE2  —  BE\     Whence,   since 
OE  =  1,  _Z><7=  ^1  —  c\ 

If  from  CA  =  1  we  subtract  DO,  we  shall  have  AD. 
That  is,  AD  =  1  —  s/\  —  c%\  butAD2  -f  DE2  =  AE\ 
and  AD'  =  2  —  2Vl  —  c2  —  c\  Adding  to  the  first 
member  of  this  last  equation  DE2,  and  to  the  second  its 
value  ca,  we  have 

AD2  -f  DB2  =  2^T^7. 

Whence,        AE  —  ^2  —  2^1  —  c\  the  value  sought. 

By  applying  this  formula  successively  to  any  known 
chord,  we  can  find  the  chord  of  one  half  the  arc,  that  of 
half  of  the  half,  and  so  on,  to  the  chords  of  the  most 
minute  arcs. 


0 

140  GEOMETRY. 

Application. 

The  greatest  chord  in  a  circle  is  its  diameter,  which  is 
2  when  the  radius  is  1 ;  therefore,  we  may  commence 
by  making  2c  =  2,  and  c  =  1. 

Then,  AE  =  ^2  —  s/T^c*  =  ^2—2^1^1  =  ^2  = 
1.41421356,  which  is  the  chord  of  90°. 
Now  make  2c  =  1.41421356,  and  c  =.70710678  =  1^2. 
We  shall  then  have, 

chord   of   45°  =  V2  —  2^t  =  V2  —  1.41421356  = 

</.58578644  =  .7653+. 

Again,  placing  2<?=.7653+,  and  applying  the  formula, 
we  would  obtain  the  chord  of  22°  30',  and  from  this  the 
chord  of  11°  15',  and  so  on,  as  far  as  we  please. 

We  may  take,  for  another  starting  point,  the  chord  of 
60°,  which  is  known  to  be  equal  to  the  radius  of  the 
circle,(Prob.  26,  B.  IV).  If,  as  above,  we  make  successive 
applications  of  the  formula,  putting  first  2c  =  1,  we  shall 
arrive  at  the  results  in  the  following 


TABLE 

• 

ion 

lof60°, 

=  \  of  a 

circumference, 

1.0000000000 

a 

"  30°, 

i  u 

—    72 

H 

.5176380902 

a 

"  15°, 

1    u 

—    HI 

ti 

.2610523842 

iC 

"  7°  30', 

_   1   u 

—     45 

U 

.1308062583 

u 

"  3°  45', 

1   U 

—    "55 

« 

.0654381655 

a 

"  1°  52'  30", 

i  a 

—  TS2 

u 

.0327234632 

u 

"  56'  15", 

-  iii  " 

a 

.0163622792 

{( 

"  28'  7"  30"', 

i  a 

—  75S 

a 

.0081812080 

a 

"  14'  3"  45"', 

1   « 

—  1535 

u 

.0040906112 

« 

"  7'  1"  52  J'", 
etc. 

1    (C 

=  3(J72 

etc. 

a 

.0020453068 

It  is  obvious  that  an  arc  so  small  as  seven  minutes  of  a 
degree  can  differ  but  very  little  from  its  chord ;  therefore, 
if  we  take  .002045307  to  be  the  true  value  of  the  3^2  of 
the  circumference,  the  whole  circumference  must  be  the 


BOOK   V.  141 

product  of  .002045307  by  3072,  which  is  6.283183104  = 
circumference  whose  radius  is  unity.  The  half  of  this, 
3.141592552,  is  the  semi-circumference,  the  more  exact 
value  of  which,  as  stated,  (Prop.  6),  is  3.141592653. 

The  value  of  the  half  circumference  being  now  deter- 
mined, if  that  of  any  arc  whatever  be  required,  we  have 
merely  to  divide  3.141592,  etc.,  by  10800,  the  number  of 
minutes  in  a  semi-circumference,  and  multiply  the  quo- 
tient by  the  number  of  minutes  in  the  arc  whose  length 
is  required. 

But  this  investigation  has  been  carried  far  enough  for 
our  present  purposes.  It  will  be  resumed  under  the 
subject  of  Trigonometry. 

We  insert  the  following  beautiful  theorem  for  the  tri- 
section  of  an  arc,  although  not  necessary  for  practical 
application.  Those  not  acquainted  with  cubic  equations 
may  omit  it. 

PROPOSITION    VIII.  — THEOREM. 

Given,  the  chord  of  any  arc,  to  determine  the  chord  of  one 
third  of  such  arc. 

Let  AE  be  the  given  chord,  and 
conceive  its  arc  divided  into  three 
equal  parts,  as  represented  by  AB, 
BD,  and  BE. 

Through  the  center  draw  BOG,  and 
draw  AB.  The  two  A's,  CAB  and 
ABE,  are  equiangular ;  for,  the  angle 
FAB,  being  at  the  circumference,  is 
measured  by  one  half  the  arc  BE,  which  is  equal  to  AB, 
and  the  angle  BOA,  being  at  the  center,  is  measured  by 
the  arc  AB ;  therefore,  the  angle  FAB  =  the  angle  BOA ; 
but  the  angle  OBA  or  EBA,  is  common  to  both  tri- 
angles ;  therefore,  the  third  angle,  OAB,  of  the  one  tri- 
angle, is  equal  to  the  third  angle,  AFB,  of  the  other, 


142  GEOMETRY. 

(Th.  12,  B.  I,  Cor.  2),  and  the  two  triangles  are  equi- 
angular and  similar. 

But  the  A  AOB  is  isosceles ;  therefore,  the  A  AFB  is 
also  isosceles,  and  AB  —  AF,  and  we  have  the  following 
proportions : 

CA  :  AB  n  AB  :  BF. 

Now,  let  AF  =  c,  AB  =  x,AC=  1.  Then  AF=  x,  and 
EF=  e  —  x,  and  the  proportion  becomes, 

1  :  x  : :  x  :  BF.     Hence,  BF=  x\ 

Also,  Fa  =  2  —  x\ 

As  AF  and  BG  are  two  chords  intersecting  each  other 
at  the  point  F,  we  have, 

GFx  FB  =  AFx  FF,  (Th.  17,  B.  III). 

That  is,  (2  —  x2)  x2  =  x(c  —  x); 

or,  x3  —  Sx  =  —  e. 

If  we  suppose  the  arc  AF  to  be  60  degrees,  then  c  =  1, 
and  the  equation  becomes  ar5  —  Sx  =  —  1;  a  cubic  equa- 
tion, easily  resolved  by  Horner's  method,  (Robinson's 
Algebra,  University  Ed.,  Art.  193),  giving  x  =  .347296  +, 
the  chord  of  20°.  This  again  may  be  taken  for  the  value 
of  c,  and  a  second  solution  will  give  the  chord  of  6°  40', 
and  so  on,  trisecting  successively  as  many  times  as  we 
please. 

PRACTICAL   PROBLEMS. 

The  theorems  and  problems  with  which  we  have  been 
thus  far  occupied,  relate  to  plane  figures;  that  is,  to 
figures  all  of  whose  parts  are  situated  in  the  same  plane. 
It  yet  remains  for  us  to  investigate  the  intersections  and 
relative  positions  of  planes ;  the  relations  and  positions 
of  lines  with  reference  to  planes  in  which  they  are  not 
contained ;  and  the  measurements,  relations,  and  proper- 
ties of  solids,  or  volumes.  But  before  we  proceed  to  this, 
it  is  deemed  advisable  to  give  some  practical  problems 
for  the  purpose  of  exercising  the  powers  of  the  student, 


*      BOOK  V.  143 

and  of  fixing  in  his  mind  those  general  geometrical  prin- 
ciples with  which  we  must  now  suppose  him  to  be 
acquainted. 

1.  The  base  of  an  isosceles  triangle  is  6,  and  the  oppo- 
site angle  is  60° ;  required  the  length  of  each  of  the  other 
two  equal  sides,  and  the  number  of  degrees  in  each  of 
the  other  angles. 

2.  One  angle  of  a  right-angled  triangle  is  30° ;  what 
is  the  other  angle  ?  Also,  the  least  side  is  12,  what  is 
the  hypotenuse  ? 

A        j  The  hypotenuse  is  24,  the  double  of  the  least 
JLm'    I      side.     Why? 

3.  The  perpendicular  distance  between  two  parallel 
lines  is  10 ;  what  angles  must  a  line  of  20  make  with 
these  parallels  to  extend  exactly  from  the  one  to  the 
other?  Arts.  The  angles  must  be  60°  and  120°. 

4.  The  perpendicular  distance  between  two  parallels 
is  20  feet,  and  a  line  is  drawn  across  them  at  an  angle  of 
45°  ;  what  is  its  length  between  the  parallels  ? 

Arts.  20^2. 

5.  -Two  parallels  are  8  feet  asunder,  and  from  a  point 
in  one  of  the  parallels  two  lines  are  drawn  to  meet  the 
other ;  the  length  of  one  of  these  lines  is  10  feet,  and 
that  of  the  other  15  feet ;  what  is  the  distance  between 
the  points  at  which  they  meet  the  other  parallel  ? 

Arts.  6.69  ft,  or  18.69  ft.     (See  Th.  39,  B.  I). 

6.  Two  parallels  are  12  feet  asunder,  and  from  a  point 
on  one  of  them  two  lines,  the  one  20  feet  and  the  other 
18  feet  in  length,  are  drawn  to  the  other  parallel ;  what 
is  the  distance  between  the  two  lines  on  the  other  parallel, 
and  what  is  the  area  of  the  triangle  so  formed  ? 

r  The  distance  on  the  other  parallel  is  29.416 
Arts.    I      feet,  or  2.584  feet;  and  the  area  of  the  tri- 
(      angle  is  176.496,  or  15.504  square  feet. 

7.  The  diameter  of  a  circle  is  12,  and  a  chord  of  the 


144  GEOMETRY. 

circle  is  4;   what  is   the  length   of  the   perpendicular 
drawn  from  the  center  to  this  chord  ?   (See  Th.  3,  B.  III). 

Arts.  W2. 

8.  Two  parallel  chords  in  a  circle  were  measured  and 
found  to  be  8  feet  each,  and  their  distance  asunder  was 
6  feet ;  what  was  the  radius  of  the  circle  ? 

Ans.  5  feet. 

9.  Two  chords  on  opposite  sides  of  the  center  of  a 
circle  are  parallel,  and  one  of  them  has  a  length  of  16 
and  the  other  of  12  feet,  the  distance  between  them 
being  14  feet.     "What  is  the  diameter  of  the  circle  ? 

Am.  20  feet. 

10.  An  isosceles  triangle  has  its  two  equal  sides,  15 
each,  and  its  base  10.  "What  must  be  the  altitude  of  a 
right-angled  triangle  on  the  same  base,  and  having  an 
equal  areaV 

11.  From  the  extremities  of  the  base  of  any  triangle, 
draw  lines  bisecting  the  other  sides ;  these  two  lines  in- 
tersecting within  the  triangle,  will  form  another  triangle 
on  the  same  base.  How  will  the  area  of  this  new  tri- 
angle compare  with  that  of  the  whole  triangle  ? 

Ans.  Their  areas  will  be  as  3  to  1. 

12.  Two  parallel  chords  on  the  same  side  of  the  center 
of  a  circle,  whose  diameter  is  32,  are  measured  and  found 
to  be,  the  one  20,  and  the  other  8.  How  far  are  they 
asunder?  Ans.  ^240  —  ^156"=  3  +. 

If  we  suppose  the  two  chords  to  be  on  opposite  sides  of  the 
center,  their  distance  apart  will  then  be  \^240  -f  \/l56  ==  15.49  + 
12.49  =  27.98. 

13.  The  longer  of  the  two  parallel  sides  of  a  trapezoid 
is  12,  the  shorter  8,  and  their  distance  asunder  5.  What 
is  the  area  of  the  trapezoid  ?  and  if  we  produce  the  two 
inclined  sides  until  they  meet,  what  will  be  the  area  of 
the  triangle  so  formed  ? 

Ans.  Area  of  trapezoid,  50 ;  area  of  triangle,  40;  area 
of  triangle  and  trapezoid,  90. 


BOOK    V. 


145 


14.  The  base  of  a  triangle  is  697,  one  of  the  sides  is 
534,  and  the  other  813.  If  a  line  he  drawn  bisecting  the 
angle  opposite  the  base,  into  what  two  parts  will  the 
bisecting  line  divide  the  base  ?     (See  Th.  25,  B.  II). 

A         (  The  greater  part  will  be  420.634 ; 
I  The  less  "  "     276.316. 

15.  Draw  three  horizontal  parallels,  making  the  dis- 
tance between  the  two  upper  parallels  7,  and  that  be- 
tween the  middle  and  lower  parallels  9  ;  then  place  be- 
tween the  upper  parallels  a  line  equal  to  10,  and  from 
the  point  in  which  it  meets  the  middle  parallel  draw  to 
the  lower  a^line  equal  to  11,  and  join  the  point  in  which 
this  last  line  meets  the  lower  parallel,  with  the  point  in 
the  upper  parallel,  from  which  the  line  10  was  drawn. 
Required  the  length  of  this  line,  and  the  area  of  the 
triangle  formed  by  it  and  the  two  lines  10  and  11. 

The  adjoining  figure 
will  illustrate.  Let  A  be 
the  point  on  the  upper 
parallel  from  which  the 
line  10  is  drawn.  Then, 
AF  =  7,  AB  =  10,  ' 
FBj=  VlOO  —  49  == 
•51.  : 

BR  =  FD  =  9,  BG 
=  11,  RC=  >/l2lZ78l 
=  \/40. 

Whence,  DC  =  «/51 
-f  v/40. 

AC2  =  (v/5l  +  \/40)2  +  (16)2;  AC  =  20.89,  Ans. 

The  area  of  the  triangle,  AB  C,  can  be  determined  by  first  find- 
ing the  area  of  the  trapezoid,  ABHD,  then  the  area  of  the  trian- 
gle, BHCj  and  from  their  sum  subtracting  the  area  of  the  triangle, 
ADC 

16.  Construct  a  triangle  on  a  base  of  400,  one  of  the 
angles  at  the  base  being  80°,  and  the  other  70° ;  and 
13  k 


Ans. 


146  GEOMETRY. 

determine  the  third  angle,  and  the  area  of  the  triangle 
thus  constructed. 

The  third  angle  is  30°,  and  as  nearly  as  our 
scale  of  equal  parts  can  determine  for  us,  the 
side  opposite  the  angle  80°  is  787,  and  that 
opposite  70°  is  740. 
The  exact  solution  of  problems  like  the  last,  except  in  a  few  par- 
ticular cases,  requires  a  knowledge  of  certain  lines  depending  on 
the  angles  of  the  triangle.  The  properties  and  values  of  these  lines 
are  investigated  in  trigonometry  •  and  as  we  are  not  yet  supposed 
to  be  acquainted  with  them,  we  must  be  content  with  the  approxi- 
mate solutions  obtained  by  the  constructions  and  measurements 
made  with  the  plane  scale. 

17.  If  we  call  the  mean  radius  of  the  earth  1,  the 
mean  distance  of  the  moon  will  be  60 ;  and  as  the  mean 
distance  of  the  sun  is  400  times  the  distance  of  the 
moon,  its  distance  will  be  400  times  60.  The  sun  and 
moon  appear  to  have  the  same  diameter;  supposing, 
then,  the  real  diameter  of  the  moon  to  be  2160  miles, 
what  must  be  that  of  the  sun  ? 


Let  E  be  the  center  of  the  earth,  M  that  of  the  moon,  and  S 
that  of  the  sun,  and  suppose  ENP  to  be  a  line  from  the  center  of 
the  earth,  touching  the  moon  and  the  sun. 

Then,  EM  :  MIST  : :  ES  :  SP; 

but  3/iVis  the  radius  of  the  moon,  and  SP  that  of  the  sun.     Mul- 
tiplying the  consequents  by  2,  the  above  proportion  becomes 
EM:  2MJST::  ES  :  2SP-, 

or  in  numbers,       60  :  2160    : :  400  X  60  :  2SP; 

whence,  2SP  =  sun's  diameter  =  864000  miles,  Ans. 

18.  In  Problem  15,  suppose  BO  to  be  drawn  on  the 
other  side  of  BIT,  what,  then,  will  be  the  value  of  A  O, 
and  what  the  area  of  the  triangle  ACB1 

Ans.   lA0=  16>021;  _ 

I  Area  of  triangle,  8^51,  very  nearly. 


BOOK.V.  147 

19.  A  man  standing  40  feet  from  a  building  which  was 
24  feet  wide,  observed  that  when  he  closed  one  eye,  the 
width  of  the  building  just  eclipsed  or  hid  from  view  90 
rods  of  fence  which  was  parallel  to  the  width  of  the 
building;  what  was  the  distance  from  the  eye  of  the 
observer  to  the  fence  ?  Ans.  2475  feet. 

20.  Taking  the  same  data  as  in  the  last  problem,  ex- 
cept that  we  will  now  suppose  the  direction  of  the  fence 
to  be  inclined  at  an  angle  of  45°  to  the  side  of  the 
building  which  we  see ;  what,  in  this  case,  must  be  the 
distance  between  the  eye  of  the  observer  and  the  remoter 
point  of  the  fence  ? 


Let  HF  be  the  width  of  the  house,  E  the  position  of  the  eye,  and 
AB  that  of  the  fence.  Draw  BD  perpendicular  to  EA  produced ; 
then,  since  the  triangle  ABB  is  right-angled  and  isosceles,  we  have 
AD  =  DB,  and  2AD2  =  AB2  =  (90)2;  BD  =  63.64  rods,  and  the 
similar  triangles  EFH  and  EDB  give  the  proportion 
HF  :  EF  : :  BD  :  ED  =  1750.1  feet; 
and  from  this  we  find 

"  EB2=ED2  +  BD2  =  (63.64  x  353)2  +  (1750.1)* 
Whence  EB  =  2040.94  -f  Ans. 

21.  In  a  right-angled  triangle,  ABO,  we  have  AB  = 
493,  AC  =  1425,  and  BC  =  1338 ;  it  is  required  to  divide 
this  triangle  into  parts  by  a  line  parallel  to  AB,  whose 
areas  are  to  each  other  as  1  is  to  3.  How  will  the  sides 
AC  and  BO  be  divided  by  this  line  ?   (See  Th.  20,  B.  II). 

Ans,  Into  equal  parts. 

22.  In  a  right-angled  triangle,  ABO,  right-angled  at 
B,  the  base  AB  is  320,  and  the  angle  A  is  60°  ;  required 
the  remaining  angle  and  the  other  sides. 

.        (  The  angle  0-  30°; 

I  AC=  640;  BC=  554.24. 


148  GEOMETRY. 

23.  A  hunter,  wishing  to  determine  his  distance  from 
a  village  in  sight,  took  a  point  and  from  it  laid  off  two 
lines  in  the  direction  of  two  steeples,  which  he  supposed 
equally  distant  from  him,  and  which  he  knew  to  he  100 
rods  asunder.  At  the  distance  of  50  feet  on  each  line 
from  the  common  point,  he  measured  the  distance  be- 
tween the  lines,  and  found  it  to  be  5  feet  8  inches.  How 
far  was  he  from  the  steeples  ? 

5  ft.  8  in. :  100  rods : :  50  ft. :  distance.         r  14,559  feet, 

or,  68:  lOOxfx  12:: 50: distance.       Ans'\      ?r  n*arl? 
2  I     3  miles. 

24.  A  person  is  in  front  of  a  building  which  he  knows 
to  be  160  feet  long,  and  he  finds  that  it  covers  10  minutes 
of  a  degree ;  that  is,  he  finds  that  the  two  lines  drawn 
from  his  eye  to  the  extremities  of  the  building  include 
an  angle  of  10  minutes.  What  is  his  distance  from  the 
building?  Ans    (    50,672  feet,  or 

•\  nearly  10  miles. 

Remark. — The  questions  of  distance,  with  which  we  are  at  present 
occupied,  depend  for  their  solution  on  the  properties  of  similar  tri- 
angles. In  the  preceding  example  we  apparently  have  but  one  tri- 
angle, but  we  have  in  fact  two ;  the  second  being  formed  by  the  dis- 
tances unity  on  the  lines  drawn  from  the  eye  of  the  observer,  and  the 
line  which  connects  the  extremities  of  these  units  of  distance.  This 
last  line  may  be  regarded  as  the  chord  of  the  arc  10  minutes  to  the 
radius  unity.  We  have  seen  that  the  length  of  the  arc  180°  to  the 
radius  1,  is  3.1415926 ;  hence  the  chord  of  1°  or  60/  is  0.017455,  and 
of  10/  it  must  be  0.0029088.     Therefore,  by  similar  triangles,  we  have 

0.0029088  :  160  : :  1  :  Ans.  =  ^^. 

25.  In  the  triangle,  ABO,  we  have  given  the  angles 
A  =  32°,  and  B  =  84°.  The  side  AB  is  produced,  and 
the  exterior  angle*  CBD  thus  formed,  is  bisected  by  the 
line  BU,  and  the  angle  A  is  also  bisected  by  the  line  AE, 
BE  and  AE  meeting  in  the  point  E.  "What  is  the  angle 
(7,  and  what  is  the  relation  between  the  angles  C  and  El 

Am.  (7=64°;  E=  J  O. 


BOOK  V.  149 

26.  Suppose  a  line  to  be  drawn  in  any  direction  be- 
tween two  parallels.  Bisect  the  two  interior  angles  thus 
formed  on  either  side  of  the  connecting  line,  and  prove 
that  the  bisecting  lines  meet  each  other  at  right  angles, 
and  that  they  are  the  sides  of  a  right-angled  triangle  of 
which  the  line  connecting  the  parallels  is  the  hypotenuse. 

27.  If  the  two  diagonals  of  a  trapezoid  be  drawn, 
show  that  two  similar  triangles  will  be  formed,  the 
parallel  sides  of  the  trapezoid  being  homologous  sides 
of  the  triangles.  What  will  be  the  relative  areas  of 
these  triangles  ? 

f  The  triangles  will  be  to  each  other 
Ans.  <      as  the  squares  on  the  parallel  sides 
(     of  the  trapezoid. 

28.  If  from  the  extremities  of  the  base  of  any  triangle, 
lines  be  drawn  to  any  point  within  the  triangle,  forming 
with  the  base  another  triangle ;  how  will  the  vertical 
angle  in  this  last  triangle  compare  with  that  in  the 
original  triangle  ? 

[  It  will  be  as  much  greater  than  the  angle 
in   the   original  triangle   as  the   sum  of 
Ans.  angles  at  the  base  of  the  new  triangle  is 

less   than  the  sum  of  those  at  the  base 
of  the  first. 

29.  The  two  parallel  sides  of  a  trapezoid  are  12  and 
20,  respectively,  and  their  perpendicular  distance  is  8. 
If  a  line  whose  length  is  14.5  be  drawn  between  the  in- 
clined sides  and  parallel  to  the  parallel  sides,  what  is  the 
area  of  the  trapezoid,  and  what  the  area  of  each  part, 
respectively,  into  which  the  trapezoid  is  divided  ? 

Area  of  the  whole,  128  square  units ; 
"     smaller  part,    33J  " 

"    larger        "       94|  " 

Dividing  line  at  the  distance  of  2J  from 
shorter  parallel  side. 

30.  If  we  assume  the  diameter  of  the  earth  to  be 
13* 


150  GEOMETRY. 

7956  miles,  and  the  eye  of  an  observer  be  40  feet  above 
the  level  of  the  sea,  how  far  distant  will  an  object  be, 
that  is  just  visible  on  the  earth's  surface.  (Employ  Th. 
18,  B.  Ill,  after  reducing  miles  to  feet.) 

Ans.  40992  feet  =  7  miles  4032  feet. 

31.  The  diameter  of  a  circle  is  4 ;  what  is  the  area  of 
the  inscribed  equilateral  triangle?  Ans.  3^3. 

32.  Three  brothers,  whose  residences  are  at  the  ver- 
tices of  a  triangular  area,  the  sides  of  which  are  severally 
10,  11,  and  12  chains,  wish  to  dig  a  well  which  shall  be 
at  the  same  distance  from  the  residence  of  each.  Deter- 
mine the  point  for  the  well,  and  its  distance  from  their 
residences. 

Remark.  —  Construct  a  triangle,  the  sides  of  which  are,  respectively, 
10,  11,  and  12.  The  sides  of  this  triangle  will  be  the  chords  of  a  cir- 
cle whose  radius  is  the  required  distance.  To  find  the  center  of  this 
circle,  bisect  either  two  of  the  sides  of  the  triangle  by  perpendiculars, 
and  their  intersection  will  be  the  center  of  the  circle,  and  the  location 
of  the  well. 

Ans,  The  well  is  distant  6.25  chains,  nearly,  from  each 
residence. 

33.  The  base  of  an  isosceles  triangle  is  12,  and  the 
equal  sides  are  20  each.  What  is  the  length  of  the  per- 
pendicular from  the  vertex  to  the  base ;  and  what  the 
area  of  the  triangle  ? 

Ans.  Perpendicular,  19.07;  area,  (19.07)  x  6. 

34.  The  hypotenuse    of  a  c 
right-angled    triangle     is    45 
inches,  and  the  difference  be- 
tween  the   two  sides  is   8.45 
inches.  Construct  the  triangle. 

Suppose  the  triangle  drawn  and 
represented  by  ABC,  DC  being  the 
difference  between  the  two  sides. 

Now,  by  inspection,  we  discover  the 
steps  to  be  taken  for  the  construc- 
tion of  the  triangle    As  AD  =  AB, 


BOOK   V.  151 

the  angle  ABB,  must  be  equal  to  the  angle  DBA,  and  each  equal 
to  45°. 

Therefore,  draw  any  line,  AC,  and  from  an  assumed  point  in  it 
as  D,  draw  BD,  making  the  angle  ABB  =  45°.     Take  from  a  • 

scale  of  equal  parts,  8.45  inches,  and  lay  them  off  from  D  to  C,  and 
with  C  as  a  center,  and  CB  =  45  inches  as  a  radius,  describe  an 
arc  cutting  BD  in  B.  Draw  CB,  and  from  B,  draw  BA  at  right 
angles  to  AC',  then  is  ABC  the  triangle  sought. 

Ans.  AB  =27.3;  A  C—  35.76,  when  carefully  constructed. 

35.  Taking  the  same  triangle  as  in  the  last  problem,  if  J%£*  **\ 
we  draw  a  line  bisecting  the  right  angle,  where  will  it  flo*K 
meet  the  hypotenuse  ? 

Ans.  19.5  from  B;  and  25.5  from  C. 

36.  The  diameters  of  the  hind  and  fore  wheels  of  a 
carriage,  are  5  and  4  feet,  respectively ;  and  their  centers 
are  6  feet  asunder.  At  what  distance  from  the  fore  wheels 
will  the  line,  passing  through  their  centers,  meet  the 
ground,  which  is  supposed  level?  Ans.  24  feet. 

37.  If  the  hypotenuse  of  a  right-angled  triangle  is  35, 
and  the  side  of  its  inscribed  square  12,  what  are  its  sides  ? 

Ans.  28  and  21. 

38.  "What  are  the  sides  of  a  right-angled  triangle 
having  the  least  hypotenuse,  in  which  if  a  square  be  in- 
scribed, its  side  will  be  12  ? 

c  The  sides  are  equal  to  24  each,  and  the 
Ans.   <      least  hypotenuse  is  double  the  diagonal 
I     of  the  square. 

39.  The  radius  of  a  circle  is  25 ;  what  is  the  area  of  a 
sector  of  50°  ? 

Remark.  — First  find  the  length  of  an  arc  of  50°  in  a  circle  whose 
radius  is  unity.  Then  25  times  that  will  be  the  length  of  an  arc  of 
the  same  number  of  degrees  in  a  circle  of  which  the  radius  is  25. 

t     ^^        io     a-          ..        3.14159269 
Length  of  arc  1°  radius  unity  = ^r — . 

u  50o      u        u     =  1-04719763  x  5> 

Area  of  sector  «  iM^i  x  125  X  f  =  54.541,  A?is. 
o  A 


152  GEOMETRY. 


BOOK  VI 


ON  THE  INTERSECTIONS  OF  PLANES,  AND  THE  REL- 
ATIVE POSITIONS  OF  PLANES  AND  OF  PLANES 
AND  LINES. 

DEFINITIONS. 

A  Plane  has  been  already  defined  to  be  a  surface,  such 
that  the  straight  line  which  joins  any  two  of  its  points 
will  lie  entirely  in  that  surface.     (Def.  9,  page  9.) 

1.  The  Intersection  or  Common  Section  of  two  planes  is 
the  line  in  which  they  meet. 

2.  A  Perpendicular  to  a  Plane  is  a  line  which  makes 
right  angles  with  every  line  drawn  in  the  plane  through 
the  point  in  which  the  perpendicular  meets  it;  and,  con- 
versely, the  plane  is  perpendicular  to  the  line.  The 
point  in  which  the  perpendicular  meets  the  plane  is 
called  the  foot  of  the  perpendicular. 

3.  A  Diedral  Angle  is  the  separation  or  divergence  of 
two  planes  proceeding  from  a  common  line,  and  is  meas- 
ured by  the  angle  included  between  two  lines  drawn 
one  in  each  plane,  perpendicular  to  their  common  sec- 
tion at  the  same  point. 

The  common  section  of  the  two  planes  is  called  the 
edge  of  the  angle,  and  the  planes  are  its  faces, 

4.  Two  Planes  are  perpendicular  to  each  other,  when  their 
diedral  angle  is  a  right  angle. 

5.  A  Straight  Line  is  parallel  to  a  plane,  when  it  will 
not  meet  the  plane,  however  far  produced. 


BOOK  VI.  153 

6.  Two  Planes  are  parallel,  when  they  will  not  intersect, 
however  far  produced  in  all  directions. 

7.  A  Solid  or  Polyedral  Angle  is  the  separation  or  diver- 
gence of  three  or  more  plane  angles,  proceeding  from  a 
common  point,  the  two  sides  of  each  of  the  plane  angles 
being  the  edges  of  diedral  angles  formed  by  these  plane 
angles. 

The  common  point  from  which  the  plane  angles  pro- 
ceed is  called  the  vertex  of  the  solid  angle,  and  the  inter- 
section "of  its  bounding  planes  are  called  its  edges. 

8.  A  Triedral  Angle  is  a  solid  angle  formed  by  three 
plane  angles. 

THEOREM   I. 

Two  straight  lines  which  intersect  each  other,  two  parallel 
straight  lines,  and  three  points  not  in  the  same  straight  line, 
will  severally  determine  the  position  of  a  plane. 

Let  AB  and  AC  be  two  lines 
intersecting  each  other  at  the 
point  A*,  then  will  these  lines 
determine  a  plane.  For,  conceive  A<^ 
a  plane  to  be  passed  through  AB, 
and  turned  about  AB  as  an  axis 

until  it  contains  the  point  0  in  the  line  AC.  The  plane, 
in  this  position,  contains  the  lines  AB  and  AC,  and  will 
contain  them  in  no  other.  Again,  let  AB  and  BE  be 
two  parallel  straight  lines,  and  take  at  pleasure  two 
points,  A  and  B,  in  the  one,  and  two  points,  D  and  E, 
in  the  other,  and  draw  AE  and  BD.  These  last  lines, 
from  what  precedes,  determine  the  position  of  a  plane 
which  contains  the  points  A,  B,  JD,  and  E.  And  again, 
if  A,  B,  and  0  be  three  points  not  in  the  same  straight 
line,  and  we  draw  the  lines  AB  and  AG,  it  follows, 
from  the  first  part  of  this  proposition,  that  these  points 
fix  the  plane. 


154 


GEOMETRY. 


Cor.  A  straight  line  and  a  point  out  of  it  determine 
the  position  of  a  plane. 

THEOREM   II. 

If  two  planes  meet  each  other,  their  common  points  will  be 
found  in,  and  form  one  straight  line. 

Let  B  and  D  be  any  two  of  the 
points  common  to  the  two  planes, 
and  join  these  points  by  the  straight 
line  BB ;  then  will  BB  contain  all 
the  points  common  to  the  two  planes, 
and  be  their  intersection.  For,  suppose  the  planes  have 
a  common  point  out  of  the  line  BB ;  then,  (Cor.  Th.  1), 
since  a  straight  line  and  a  point  out  of  it  determine  a 
plane,  there  would  be  two  planes  determined  by  this  one 
line  and  single  point  out  of  it,  which  is  absurd.  Hence 
the  common  section  of  two  planes  is  a  straight  line. 

Remark. — The  truth  of  this  proposition  is  implicitly  assumed  in  the 
definitions  of  this  Book. 


THEOREM    III. 

If  a  straight  line  stand  at  right  angles  to  each  of  two  other 
straight  lines  at  their  point  of  intersection,  it  will  be  at  right 
angles  to  the  plane  of  those  lines. 

Let  AB  stand  at  right  angles  to  EF&ndi 
CB,  at  their  point  of  intersection  A.  Then 
AB  will  be  at  right  angles  to  any  other 
line  drawn  through  A  in  the  plane,  pass- 
ing through  EF,  OB,  and,  of  course,  at 
right  angles  to  the  plane  itself.     (Def.  2.) 

Through  A,  draw  any  line,  A  G,  in  the 
plane  EF,  CB,  and  from  any  point  G,  draw  GH  parallel 
to  AB.  Take  HF  =  AH,  and  join  F  and  G  and  produce 
FG  to  B.     Because  HG  is  parallel  to  AB,  we  have 

FH  :  HA  ::  FG  :  GB. 


BOOK  VI.  155 

But,  in  this  proportion,  the  first  couplet  is  a  ratio  of 
equality;  therefore  the  last  couplet  is  also  a  ratio  of 
equality, 

That  is,  FG  =  GB,  or  the  line  FB  is  bisected  in  ff. 

Draw  BB,  BG,  and  BF. 

Now,  in  the  triangle  AFB,  as  the  base  FB  is  bisected 
in  G,  we  have, 

JJF2 +~AB2  =  2AG2  +  2GF2    (l)    (Th.  42,  B.  I). 

Also,  as  BF  is  the  base  of  the  A  BBF,  we  have  by  the 
same  theorem, 

~BF2  +~~BB2  =  2BG2  +  2GF2  (2) 


By  subtracting  ( 1 )  from  (2 ),  and  observing  that  BF  — 


AF  =  AB  ?  because  BAF  is  a  right  angle ;  and  BB  — 
AB2  —  AB2,  because  BAB  is  a  right  angle,  we  shall  have, 

~AB*  +~AB2  =  2BG2  —  2  AG2. 

Dividing  by  2,  and  transposing  AG-2,  and  we  have, 

AB2  +  AG2  =  BG\ 

This  last  equation  shows  that  BA  G  is  a  right  angle. 
But  AG  is  any  line  drawn  through  A,  in  the  plane  FF, 
OB ;  therefore  AB  is  at  right  angles  to  any  line  in  the 
plane,  and,  of  course,  at  right  angles  to  the  plane  itself. 

Cor.  1.  The  perpendicular  BA  is  shorter  than  any  of 
the  oblique  lines  BF,  BG,  or  BB,  drawn  from  the  point 
B  to  the  plane ;  hence  it  is  the  shortest  distance  from  a 
point  to  a  plane. 

Cor.  2.  But  one  perpendicular  can  be  erected  to  a  plane 
from  a  given  point  in  the  plane;  for,  if  there  could  be 
two,  the  plane  of  these  perpendiculars  would  intersect 
the  given  plane  in  some  line,  as  AG,  and  both  the  per- 
pendiculars would  be  at  right  angles  to  this  intersection 
at  the  same  point,  which  is  impossible. 

Cor.  8.  But  one  perpendicular  can  be  let  fall  from  a 
given  point  out  of  a  plane  on  the  plane ;  for,  if  there  can 


156  GEOMETRY. 

be  two,  let  BGr  and  BA  be  such  perpendiculars,  then 
would  the  triangle  BAGr  be  right  angled  at  both  A  and 
6r,  which  is  impossible. 

THEOREM   IV. 

If  from  any  point  of  a  perpendicular  to  a  plane,  oblique 
lines  be  drawn  to  different  points  in  the  plane,  those  oblique 
lines  which  meet  the  plane  at  equal  distances  from  the  foot  of 
the  perpendicular  are  equal;  and  those  which  meet  the  plane 
at  unequal  distances  from  the  foot  of  the  perpendicular  are 
unequal,  the  greater  distances  corresponding  to  the  longer 
oblique  lines. 

Take  any  point  B  in 
the  perpendicular  BA  to 
the  plane  ST,  and  draw 
the  oblique  lines  BO, 
BD,  and  BE,  the  points 
0,  B,  and  E,  being  equally 
distant  from  A,  the  foot 
of  the  perpendicular. 
Produce  AE  to  F,  and 
draw  BF;  then  will  BC=  BD  =  BE,  and  BF>  BE. 

For,  the  triangles  BAG,  BAD,  and  BAE  are  all  right- 
angled  at  A,  the  side  BA  is  common,  and  AC= AD=  AE 
by  construction,  hence,  (Th.  23,  B.I),  BC=BB  =  BE. 
Moreover,  since  AF^>  AE,  the  oblique  line  BF^>  BE. 

Cor.  If  any  number  of  equal  oblique  lines  be  drawn 
from  the  point  B  to  the  plane,  they  will  all  meet  the 
plane  in  the  circumference  of  a  circle  having  the  foot  of 
the  perpendicular  for  its  center.  It  follows  from  this, 
that,  if  three  points  be  taken  in  a  plane  equally  distant 
from  a  point  out  of  it,  the  center  of  the  circumference 
passing  through  these  three  points  will  be  the  foot  of  the 
perpendicular  drawn  from  the  point  to  the  plane. 


BOOK    VI.  157 

THEOREM  V. 

The  line  which  joins  any  point  of  a  perpendicular  to  a 
plane,  with  the  point  in  which  a  line  in  the  plane  is  inter- 
sected, at  right  angles,  by  a  line  through  the  foot  of  the  per- 
pendicular, will  be  at  right  angles  to  the  line  in  the  plane. 

Let  AB  be  perpendic- 
ular to  the  plane  ST,  and 
AB  a  line  through  its  foot 
at  right  angles  to  EF,  a  line 
in  the  plane.  Connect  B 
with  any  point,  as  B,  of  the 
perpendicular;  and  BD  will 
be  perpendicular  to  EF. 

Make  BF=  DE,  and  join  B  to  the  points  E,  D,  and 
F.  Since  BE  =  BF,  and  the  angles  at  B  are  right 
angles,  the  oblique  lines,  AE  and  AF,  are  equal ;  and, 
since  AE  =  AF,  we  have,  (Th.  4),  BE  =  BF;  therefore 
the  line  BB  has  its  two  points,  B  and  B,  equally  distant 
from  the  extremities  E  and  F  of  the  line  EF,  and  hence 
BB  is  perpendicular  to  EF  at  its  middle  point  B. 

Cor.  Since  FB  is  perpendicular  to  the  two  lines  AB 
and  BB  at  their  intersection,  it  is  perpendicular  to  their 
plane  ABB,  (Th.  3). 

Scholium.  —  The  inclination  of  a  line  to  a  plane  is  measured  by  the 
angle  included  between  the  given  line  and  the  line  which  joins  the 
point  in  which  it  meets  the  plane  and  the  foot  of  the  perpendicular 
drawn  from  any  point  of  the  line  to  the  plane ;  thus,  the  angle  BFA  is 
the  inclination  of  the  line  BF  to  the  plane  ST. 

THEOREM   VI. 

If  either  of  two  parallels  is  perpendicular  to  a  plane,  the 
other  is  also  perpendicular  to  the  plane. 

Let  BA  and  ED  be  two  parallels,  of  which  one,  BA, 
is  perpendicular  to  the  plane  ST;  then  will  the  other  also 
be  perpendicular  to  the  same  plane. 
14 


158 


GEOMETKY. 


The  two  parallels  de- 
termine a  plane  which 
intersects  the  given  plane 
in  AB ;  through  B  draw 
MJV  perpendicular  to 
AB;  then,  (Cor.,  Th.  5,) 
will  MJSf  be  perpendicu- 
lar to  the  plane  BAB, 
and  the  angle  MBE  is 

therefore  a  right  angle ;  but  EBA  is  also  a  right  angle, 
since  BA  and  ED  are  parallel,  and  BAB  is  a  right  angle 
by  hypothesis;  hence,  EB  is  perpendicular  to  the  two 
lines  MB  and  AB  in  the  plane  ST;  it  is  therefore  perpen- 
dicular to  the  plane,  (Th.  3). 

Cor.  1.  The  converse  of  this  proposition  is  also  true , 
that  is,  if  two  straight  lines  are  both  perpendicular  to  the  same 
plane,  the  lines  are  parallel. 

For,  suppose  BA  and  EB  to  be  two  perpendiculars ;  if 
not  parallel,  draw  through  B  a  parallel  to  BA,  and  this 
last  line  will  be  perpendicular  to  the  plane ;  but  EB  is 
a  perpendicular  by  hypothesis,  and  we  should  have  two 
perpendiculars  erected  to  the  plane  at  the  same  point, 
which  is  impossible,  (Cor.  2,  Th.  3). 

Cor.  2.  If  two  lines  lying  in  the  same  plane  are  each 
parallel  to  a  third  line  not  in  the  same  plane,  the  two 
lines  are  parallel.  For,  pass  a  plane  perpendicular  to 
the  third  line,  and  it  will  be  perpendicular  to  each  of  the 
others;  hence  they  are  parallel, 


THEOREM   VII. 


A  straight  line  is  parallel  to  a  plane,  when  it  is  parallel 
to  a  line  in  the  plane. 

Suppose  the  line  MN  to  be  parallel  to  the  line  CB,  in 
the  plane  ST;  then  will  ifefJVbe  parallel  to  the  plane  ST 


BOOK    VI.  159 

For,  CB  being  in  the  plane       

ST,   and    at    the    same    time     g 

parallel  to  MN,  it  must  be  the 

intersection   of  the  plane   of 

these  parallels  with  the  plane 

ST;   hence,  if  MN  meet  the 

plane  ST,  it  must  do  so  in  the  T 

line  CB,  or  OB  produced ;  but  MN  and  CB  are  parallel, 

and  cannot  meet;  therefore  MN,  however  far  produced, 

can  have  no  point  in  the  plane  ST,  and  hence,  (Def.  5),  it 

is  parallel  to  this  plane. 

THEOREM    VIII. 

If  two  lines  are  parallel,  they  will  be  equally  inclined  to 
any  given  plane. 
Let  AB  and  CB  be 

•n  tv 

two  parallels,   and   ST 

any  plane  met  by  them 

in    the    points    A    and      V 

'C;   then  will  the  lines       \ 

AB  and  CB  be  equally        \      ^         "~c  V 

inclined   to    the    plane         \ \ 

ST.  T 

For,  take  any  distance,  AB,  on  one  of  these  parallels, 
and  make  CB  =  AB,  and  draw  A  C  and  BB.  From  the 
points  B  and  B  let  fall  the  perpendiculars,  BB  and  BF, 
on  the  plane ;  join  their  feet  by  the  line  EF,  and  draw 
AE  and  OF. 

Now,  since  AB  is  equal  and  parallel  to  CB,  ABB  Cm 
a  parallelogram,  and  BB  is  equal  and  parallel  to  A  0, 
and  BB  is  parallel  to  the  plane  ST,  (Th.  7) ;  and,  since 
BE  and  BF  are  both  perpendicular  to  this  plane,  they 
are  parallel ;  but  BB  and  EF  are  in  the  plane  of  these 
parallels;  and  as  EF  is  in  the  plane  ST,  and  BB  is 
parallel  to  this  plane,  these  two  lines  must  be  parallel 
and  equal,  and  BBFE  is  also  a  parallelogram.     Now, 


160  GEOMETRY. 

we  have  shown  that  BD  is  equal  and  parallel  to  AC,  and 
EF  equal  and  parallel  to  BD;  hence,  (Cor.  2,  Th.  6), 
EFw  equal  and  parallel  to  AC,  and  ACFE  is  a  parallel- 
ogram, and  AE  =  CF.  The  triangles  ABE  and  CDF 
have,  then,  the  sides  of  the  one  equal  to  the  sides  of  the 
other,  each  to  each,  and  their  angles  are  consequently 
equal;  that  is,  the  angle  BAE  is  equal  to  the  angle 
DCF;  but  these  angles  measure  the  inclination  of  the 
lines  AB  and  CD  to  the  plane  ST,  (Scholium,  Th.  5). 

Scholium.  —  The  converse  of  this  proposition  is  not  generally  true  ; 
that  is,  straight  lines  equally  inclined  to  the  same  plane  are  not  neces- 
sarily parallel. 

THEOREM  IX. 

The  intersections  of  two  parallel  planes  by  a  third  plane, 
are  parallel. 

Let  the  planes  QR  and  ST  be  intersected  by  the  third 
plane,  AD :  then  will  the  intersections,  AB  and  CD,  be 
parallel. 

Since  the  lines  AB  and  CD  are  in  the  same  plane,  if 
they  are  not  parallel,  they  will 
meet  if  sufficiently  produced; 
but  they  cannot  meet  out  of  the 
planes  QR  and  ST,  in  which 
they  are  respectively  found; 
therefore,  any  point  common  to 
the  lines,  must  be  at  the  same 
time  common  to  the  planes ;  and 
since  the  planes  are  parallel, 
they  have  no  common  point,  and  the  lines,  therefore,  do 
not  intersect ;  hence  they  are  parallel. 

THEOREM    X. 

If  two  planes  are  perpendicular  to  the  same  straight  line, 
they  are  parallel  to  each  other. 

Let  QR  and  ST  be  two  planes,  perpendicular  to  the 
line  ABm,  then  will  these  planes  be  parallel* 


BOOK  VI.  161 

For,  if  not  parallel,  suppose  M  to  be  a  point  in  their 
line  of  intersection,  and  n 

from    this  point    draw 

lines  to  the  extremities  

of    the     perpendicular    m 

AB,  thus  forming  a  tri-  "^oii. 

angle,     MAB.      .Now, 

since   the    line  AB    is  \ \ 

T 
perpendicular   to    both 

planes,  it  is  perpendicular  to  each  of  the  lines  MA  and 
MB,  drawn  through  its  feet  in  the  planes,  (Def.  2) ; 
hence,  the  triangle  has  two  right  angles,  which  is  impos- 
sible; the  planes  cannot  therefore  meet  in  any  point  as 
My  and  are  consequently  parallel. 

Cor.  Conversely :  The  straight  line  which  is  perpendicu- 
lar to  one  of  the  parallel  planes,  is  also  perpendicular  to  the 
other.  For,  if  AB  be  perpendicular  to  the  plane  QB, 
draw  in  the  other  plane,  through  the  point  in  which  the 
perpendicular  meets  it,  any  line,  as  AC.  The  plane  of 
tjie  lines  AB  and  A  0  will  intersect  the  plane  QR  in  the 
line  BD  ;  and  since  the  planes  are  parallel  by  hypothesis, 
the  lines  A  O  and  BD  must  be  parallel,  (Th.  9) ;  but  the 
angle  DBA  is  a  right  angle ;  hence,  BAG  must  be  a  right 
angle,  and  the  line  BA  is  perpendicular  to  any  line  what- 
ever drawn  in  the  plane  through  the  point  A ;  BA  is 
therefore  perpendicular  to  the  plane  ST. 


THEOREM   XI. 

If  two  straight  lines  be  drawn  in  any  direction  through 
parallel  planes,  the  planes  will  cut  the  lines  proportionally. 

Conceive  three  planes  to  be  parallel,  as  represented 
in  the  figure,  and  take  any  points,  A  and  B,  in  the  first 
and  third  planes,  and  draw  AB,  the  line  passing  through 
the  second  plane  at  E. 

14*  L 


162 


GEOMETRY. 


G\ 


Also,  take  any  other  two  points,  as 
O  and  D,  in  the  first  and  third  planes, 
and  draw  OB,  the  line  passing  through 
the  second  plane  at  F. 

Join  the  two  lines  by  the  diagonal 
AB,  which  passes  through  the  second 
plane  at  #.     Draw  BB,  EQ,  'OJFi  and 
A  0.     We  are  now  to  prove  that, 
AE  :  EB  ::  OF  :  FB. 

For  the  sake  of  perspicuity,  put  AGr  =  X,  and  GB=Y. 

As  the  planes  are  parallel,  BB  is  parallel  EGr ;  then,  in 
the  two  triangles  ABB  and  AEGr,  we  have,  (Th.  17, 
B.H); 

AE  :  EB  : :  X  :  Y. 

Also,  as  the  planes  are  parallel,  GrF  is  parallel  to  A  Cy 
and  we  have, 

OF  :  FB  : :  X  :   Y. 

By  comparing  the  proportions,  and  applying  Th.  6, 
B.  II,  we  have 

AE  :  EB  n  OF  i  FB. 


THEOREM    XII, 


If  a  straight  line  is  perpendicular  to  a  plane,  all  planes 
passing  through  that  line  will  be  perpendicular  to  the  plane. 

Let  MNhe  a  plane,  and  AB  a  per- 
pendicular to  it.  Let  BO  be  any 
other  plane,  passing  through  AB ; 
this  plane  will  be  perpendicular  to 

mjst. 

Let  BB  be  the  common  intersec- 
tion of  the  two  planes,  and  from 
the  point  B,  draw  BE  at  right  angles  to  BB. 

Then,  as  AB  is  perpendicular  to  the  plane  MJSf,  it  is 
perpendicular  to  every  line  in  that  plane,  passing  through 


BOOK    VI. 


163 


B;  (Def.  2,) ;  therefore,  ABE  is  a  right  angle.  But  the 
angle  ABU,  (Def.  3),  measures  the  inclination  of  the  two 
planes ;  therefore,  the  plane  OB  is  perpendicular  to  the 
plane  MN\  and  thus  we  can  show  that  any  other  plane, 
passing  through  AB,  will  be  perpendicular  to  JOT. 
Hence  the  theorem. 


THEOREM   XIII. 


If  two  planes  are  perpendicular  to  each  other,  and  a  line 
be  drawn  in  one  of  them  perpendicular  to  their  common  in- 
tersection, it  will  be  perpendicular  to  the  other  plane. 

Let  the  two  planes,  QB  and  ST,  be  perpendicular  to 
each  other,  and  draw  in  QB  the  line  CD  at  right  angles 
to  their  common  intersection,  B  V;  then  will  this  line  be 
perpendicular  to  the  plane  ST. 

In  the  plane  iSTdraw  ED,  perpen- 
dicular to  VB  at  the  point  D. 
Then,  since  the  planes  QB  and  ST 
are  perpendicular  to  each  other,  the 
angle  ODE  is  a  right  angle,  and 
CD  is  perpendicular  to  the  two 
lines,  ED  and  VB,  passing  through 
its  foot  in  the  plane  ST.  CD  is  therefore  perpendicular 
to  the  plane  ST,  (Th.  3). 

Cor.  Conversely:  if  we  erect  a  perpendicular  to  the 
plane  ST,  at  any  point,  D,  of  its  intersection  with  the 
plane  QB,  this  perpendicular  will  lie  in  the  plane  QB. 
For,  if  it  be  not  in  this  plane,  we  can  draw  in  the  plane 
the  line  CD,  at  right  angles  to  VB ;  and,  from  what  has 
been  shown  above,  CD  is  perpendicular  to  the  plane  ST, 
and  we  should  thus  have  two  perpendiculars  erected  to 
the  plane,  ST,  at  the  same  point,  which  is  impossible, 
(Cor.  2,  Th.  2>\ 


164 


GEOMETRY. 


THEOREM   XIV. 


The  common  intersection  of  two  planes,  loth  of  which  are 
perpendicular  to  a  third  plane,  will  also  be  perpendicular  to 
the  third  plane. 

Let  MN  be  the  common 
intersection  of  the  two 
planes,  QR  and  VX,  both 
of  which  are  perpendicular 
to  the  plane  ST;  then  will 
MJSfbe  perpendicular  to  the 
plane  ST.  For,  if  we  erect 
a  perpendicular  to  the  plane 
ST,  at  the  point  M,  it  will 
lie  in  both  planes   at  the 

same  time,  (Cor.  Th.  13);  and  this  perpendicular  must 
therefore  be  their  intersection.     Hence  the  theorem. 


THEOREM   XV. 

Parallel  straight  lines  included  between  parallel  planes, 
are  equal. 

Let  J. B  and  D  0  be  two  parallel  lines, 
included  by  the  two  parallel  planes, 
QR  and  ST;  then  will  AB  =  BO. 

For,  the  plane  A  0,  of  the  parallel 
lines,  intersects  the  planes,  QR  and  ST, 
in   the   parallel  lines,   AB  and  BO, 
(Th.  9) ;    hence  ABBO  is  a  parallelogram,  and  its  oppo- 
site sides,  AB  and  BO,  are  equal. 

Oor.  It  follows  from  this  proposition,  that  parallel  planes 
are  everywhere  equally  distant ;  for,  two  perpendiculars 
drawn  at  pleasure  between  the  two  planes  are  parallel 
lines,  (Cor.  1,  Th.  6),  and  hence  are  equal ;  but  these  per- 
pendiculars measure  the  distance  between  the  planes. 


V 

1\ 

s   . 

R 

V 

^^ 

-\ 

BOOK    VI. 


165 


THEOREM    XVI. 

Two  planes  are  parallel  when  two  lines  not  parallel,  lying 
in  the  one,  are  respectively  parallel  to  two  lines  lying  in  the 
other. 

Let  QR  and  ST  be 
two  planes,  the  first 
containing  the  two 
lines  AB  and  CD 
which  intersect  each 
other  at  U,  and  the 
second  the  two  lines 
LM  and  NO,  respect- 
ively parallel  to  AB 
and  OB;  then  will 
these  planes  be  par- 
allel. 

For,  if  the  two  planes 
are  not  parallel,  they  must  intersect  when  sufficiently 
produced;  and  their  common  section  lying  in  both  planes 
at  the  same  time,  would  be  a  line  of  the  plane  QR.  Now, 
the  lines  AB  and  OB  intersect  each  other  by  hypothesis ; 
hence  one  or  both  of  them  must  meet  the  common  sec- 
tion of  the  two  planes.  Suppose  AB  to  meet  this  com- 
mon section ;  then,  since  AB  and  LM  are  parallel,  they 
determine  a  plane,  and  AB  cannot  meet  the  plane  ST  in 
a  point  out  of  the  line  LM ;  but  AB  and  LM  being  par- 
allel, have  no  common  point.  Hence,  neither  AB  nor 
OB  can  meet  the  common  section  of  the  two  planes ;  that 
is,  they  have  no  common  section,  and  are  therefore  par- 
allel. 

Oor.  Since  two  lines  which  intersect  each  other,  deter- 
mine a  plane,  it  follows  from  this  proposition,  that  the 
plane  of  two  intersecting  lines  is  parallel  to  the  plane  of  two 
other  intersecting  lines  respectively  parallel  to  the  first  lines. 


166 


GEOMETRY. 


THEOREM   XVII. 


W7ien  two  intersecting  lines  are  respectively  parallel  to  two 
other  intersecting  lines  lying  in  a  different  plane,  the  angles 
formed  by  the  last  two  lines  will  be  equal  to  those  formed  by 
the  first  two,  each  to  each,  and  the  planes  of  the  angles  will  be 
parallel. 

Let  QR  be  the  plane 
of  the  two  lines  AB 
and  CD,  which  inter- 
sect each  other  at  the 
point  E,  and  ST  the 
plane  of  the  two  lines 
LM  and  NO,  respect- 
ively parallel  to  AB 
and  CD ;  then  will  the 
[_BED  -  \__MPO, 
and  L  BEQ  =  L 
MPN,  etc.,  and  the 
planes  QR  and  ST 
will  be  parallel. 

That  the  plane  of  one  set  of  angles  is  parallel  to  that 
of  the  other,  follows  from  the  Corollary  to  Theorem  16  ; 
we  have  then  only  to  show  that  the  angles  are  equal, 
each  to  each. 

Take  any  points,  B  and  D,  on  the  lines  AB  and  CD, 
and  draw  BD.  Lay  off  PM,  equal  to  and  in  the  same 
direction  with  EB,  and  PO,  equal  to  and  in  the  same 
direction  with  ED,  and  draw  MO.  Kow,  since  the  planes 
QR  and  ST  are  parallel,  and  ED  is  equal  and  parallel  to 
PO,  ED  OP  is  a  parallelogram,  and  DO  is  equal  and  par- 
allel to  EP.  For  the  same  reason,  BM  is  equal  and 
parallel  to  EP;  therefore,  BDOM  is  a  parallelogram,  and 
MO  is  equal  and  parallel  to  BD.  Hence  the  A's,  EBD 
and  PMO,  have  the  sides  of  the  one  equal  to  the  sides 
of  the  other,  each  to  each ;  they  are  therefore  equal,  and 


BOOK  VI.  167 

the  [_MPO  =  the  \_BED.  In  the  same  manner  it  can 
be  proved  that  [_BEC  =  [_MPJST,  etc. 

Cor.  1.  The  plane  of  the  parallels  AB  and  LM  is  in- 
tersected by  the  plane  of  the  parallels  CD  and  NO,  in  the 
line  EP.  Now,  EB  and  ED  are  the  intersections  of  these 
two  planes  with  the  plane  QB,  and  PM  and  PO  are  the 
intersections  of  the  same  planes  with  the  parallel  plane 
ST.  It  has  just  been  proved  that  the  \_  BED  =  [_MPO. 
Hence,  if  the  diedral  angle  formed  by  two  planes,  be  cut  by 
two  parallel  planes,  the  intersections  of  the  faces  of  the  diedral 
angle  with  one  of  these  planes  will  include  an  angle  equal 
to  that  included  by  the  intersections  of  the  faces  with  the  other 
plane. 

Cor.  2.  The  opposite  triangles  formed  by  joining  the  cor- 
responding extremities  of  three  equal  and  parallel  straight 
lines  lying  in  different  planes,  will  be  equal  and  the  planes  of 
the  triangles  will  be  parallel. 

Let  EP,  BM,  and  DO,  be  three  equal  and  parallel 
straight  lines  lying  in  different  planes.  By  joining  their 
corresponding  extremities,  we  have  the  triangles  EBD 
and  PMO.  Now,  since  EP  and  BM  are  equal  and 
parallel,  EBMP  is  a  parallelogram,  and  EB  is  equal  and 
parallel  to  PM;  in  the  same  manner,  we  show  that  ED 
is  equal  and  parallel  to  PO,  and  BD  to  MO',  hence  the 
triangles  are  equal,  having  the  three  sides  of  the  one, 
respectively,  equal  to  the  three  sides  of  the  other. 
That  their  planes  are  parallel,  follows  from  Cor.,  Theo- 
rem 16. 

THEOREM   XVIII. 

Any  one  of  the  three  plane  angles  bounding  a  triedral 
angle,  is  less  than  the  sum  of  the  other  two. 

Let  A  be  the  vertex  of  a  solid  angle,  bounded  by  the 
three  plane  angles,  BAC,  BAD,  and  DAC;  then  will  any 
one  of  these  three  angles  be  less  than  the  sum  of  the 


168  GEOMETRY. 

other  two.     To  establish  this  proposition,  we  have  only 
to  compare  the  greatest  of  the  three 
angles  with  the  sum  of  the  other 
two. 

Suppose,  then,  BAC  to  be  the 
greatest  angle,  and  draw  in  its  plane  B4 
the   line  AE,  making    the    angle 
CAB  equal  to  the  angle  CAD,     On  1) 

AH,  take  any  point,  E,  and  through  it  draw  the  line  CUB. 
Take  AD,  equal  to  AE,  and  draw  BD  and  DC. 

Now,  the  two  triangles,  CAD  and  CAE,  having  two 
sides  and  the  included  angle  of  the  one  equal  to  the  two 
sides  and  included  angle  of  the  other,  each  to  each,  are 
equal,  and  CE  =  CD',  but  in  the  triangle,  BDC,  BC<i 
BD  +  DC.  Taking  EC  from  the  first  member  of  this 
inequality,  and  its  equal,  DC,  from  the  second,  we  have, 
BE  <  BD.  In  the  triangles,  BAE  and  BAD,  BA  is 
common,  and  AE  =  AD  by  construction ;  but  the  third 
side,  BD,  in  the  one,  is  greater  than  the  third  side,  BE, 
in  the  other ;  hence,  the  angle  BAD  is  greater  than  the 
angle  BAE,  (Th.  22,  B.  I) ;  that  is,  [_BAE  <  [_BAD; 
adding  the  \_EAC  to  the  first  member  of  this  inequality, 
and  its  equal,  the  [_DAC,  to  the  other,  we  have 

l_BAE+  [_EAC<  \_BAD  +  \_DAC. 
And,  as  the  l_BAC  is  made  up  of  the  angles  BAE  and 
EAC,  we  have,  as  enunciated, 

[_BAC<:  [_BAD  +  [_DAC. 

THEOREM    XIX. 

The  sum  of  the  plane  angles  forming  any  solid  angle,  is 
always  less  than  four  right  angles. 

Let  the  planes  which  form  the  solid  angle  at  A,  be  cut 
by  another  plane,  which  we  may  call  the  plane  of  the 
base,  BCDE.  Take  any  point,  a,  in  this  plane,  and  draw 
aB,  aC,  aD,  aE,  etc.,  thus  making  as  many  triangles  on 


BOOK    VI. 


169 


the  plane  of  the  base  as  there  are  tri- 
angular planes  forming  the  solid  angle 
A.  Now,  since  the  sum  of  the  angles 
of  every  A  is  two  right  angles,  the  sum 
of  all  the  angles  of  the  A's  which 
have  their  vertex  in  A,  is  equal  to  the 
sum  of  all  angles  of  the  A's  which  have 
their  vertex  in  a.  But,  the  angles  BOA 
+  AOB,  are,  together,  greater  than 
the  angles  BOa  +  aOD,  or  BCD,  by  the  last  proposition. 
That  is,  the  sum  of  all  the  angles  at  the  bases  of  the  A's 
which  have  their  vertex  in  A,  is  greater  than  the  sum  of 
all  the  angles  at  the  bases  of  the  A's  which  have  their 
vertex  in  a.  Therefore,  the  sum  of  all  the  angles  at  a  is 
greater  than  the  sum  of  all  the  angles  at  A ;  but  the  sum 
of  all  the  angles  at  a  is  equal  to  four  right  angles ;  there- 
fore, the  sum  of  all  the  angles  at  A  is  less  than  four  right 
angles. 

THEOREM    XX." 

If  two  solid  angles  are  formed  by  three  plane  angles  respect- 
ivety  equal  to  each  other,  the  planes  which  contain  the  equal 
angles  will  be  equally  inclined  to  each  other. 

Let  the  [__ASO=t\iQ[_DTF, 
the  [_ASB  =  the  [_BTE,  and 
the  [_BSO=  the  [_FTF;  then 
will  the  inclination  of  the 
planes,  ASO,  ASB,  be  equal 
to  that  of  the  planes,  DTF, 
DTF. 

Having  taken  SB  at  pleas- 
ure, draw  BO  perpendicular 

to  the  plane  ASO;  from  the  point  0,  at  which  that  perpen- 
dicular meets  the  plane,  draw  OA  and  00,  perpendicular 
to  SA  and  SO;  draw  AB  and  BO;  next  take  TF  =  SB, 
and  draw  FP  perpendicular  to  the  plane  DTF;  from  the 
15 


170  GEOMETRY. 

point  P,  draw  PB  and  PF,  perpendicular  to  TB  and 
TF;  lastly,  draw  BE  and  FF. 

The  triangle  SAB,  is  right-angled  at  A,  and  the  tri- 
angle TBE,  at  B,  (Th.  5) ;  and  since  the  [__  ^£5  =  the 
L  B  TF,  we  have  |_  SB  A  -  L  ^-#0 ;  likewise,  SB=TE; 
therefore,  the  triangle  $Ai?  is  equal  to  the  triangle  TBF; 
hence,  SA  =  TB,  and  AB  =  Z>i?.  In  like  manner  it 
may  be  shown  that  SO  =  TF,  and  BO  =  FF.  That 
granted,  the  quadrilateral  SAOO  is  equal  to  the  quadri- 
lateral TBPF;  for,  place  the  angle  ASQ  upon  its  equal, 
BTF,  and  because  SA  =  2rD,  and  £#  =  TF,  the  point  J. 
will  fall  on  B,  and  the  point  0  on  jP;  and,  at  the  same  time, 
A  0,  which  is  perpendicular  to  SA,  will  fall  on  PB,  which 
is  perpendicular  to  TB,  and,  in  like  manner,  00  on  PF; 
wherefore,  the  point  0  will  fall  on  the  point  P,  and  A  0 
will  be  equal  to  DP.  But  the  triangles,  AOB,  DPE,  are 
right  angled  at  0  and  P ;  the  hypotenuse  AB  ==  BE,  and 
the  side  AO  =  BP;  hence,  those  triangles  are  equal, 
(Cor,  Th.  39,  B.  I),  and  [_A0B=[_PBE.  The  angle  OAB 
is  the  inclination  of  the  two  planes,  ASB,  ASO;  the  angle 
PBE  is  that  of  the  two  planes,  BTE,  BTF;  conse- 
quently, those  two  inclinations  are  equal  to  each  other. 

Hence  the  theorem. 

Scholium  1.  —  The  angles  which  form  the  solid  angles  at  S  and  T, 
may  be  of  such  relative  magnitudes,  that  the  perpendiculars,  BO  and 
EP,  may  not  fall  within  the  bases,  ASC  and  BTF;  but  they  will 
always  either  fall  on  the  bases,  or  on  the  planes  of  the  bases  produced, 
and  0  will  have  the  same  relative  situation  to  Ay  S,  and  C,  as  P  has 
to  D,  T,  and  jF.  In  case  that  0  and  P  fall  on  the  planes  of  the  bases 
produced,  the  angles  BCO  and  EFP,  would  be  obtuse  angles ;  but  the 
demonstration  of  the  problem  would  not  be  varied  in  the  least. 

Scholium  2.  —  If  the  plane  angles  bounding  one  of  the  triedral 
angles  be  equal  to  those  of  the  other,  each  to  each,  and  also  be  simi- 
larly arranged  about  the  triedral  angles,  these  solid  angles  will  be  ab- 
solutely equal.  For  it  was  shown,  in  the  course  of  the  above  demon- 
stration, that  the  quadrilaterals,  SA OC  and  TDPF,  were  equal;  and 
on  being  applied,  the  point  0  falls  on  the  point  P;  and  since  the  trian- 
gles A  OB  and  DPE  are  equal,  the  perpendiculars  OB  and  PE  are 


BOOK  VI.  171 

also  equal.  Now,  because  the  plane  angles  are  like  arranged  about 
the  triedral  angles,  these  perpendiculars  lie  in  the  same  direction ; 
hence  the  point  B  will  fall  on  the  point  E,  and  the  solid  angles 
will  exactly  coincide. 

Scholium  3. — When  the  planes  of  the  equal  angles  are  not  like  dis- 
posed about  the  triedral  angles,  it  would  not  be  possible  to  make  these 
triedral  angles  coincide ;  and  still  it  would  be  true  that-the  planes  of 
the  equal  angles  are  equally  inclined  to  each  other.  Hence,  these 
triedral  angles  have  the  plane  and  diedral  angles  of  the  one,  equal  to 
the  plane  and  diedral  angles  of  the  other,  each  to  each,  without  having 
of  themselves  that  absolute  equality  which  admits  of  superposition. 
Magnitudes  which  are  thus  equal  in  all  their  component  parts,  but 
will  not  coincide,  when  applied  the  one  to  the  other,  are  said  to  be 
symmetrically  equal.  Thus,  two  triedral  angles,  bounded  by  plane 
angles  equal  each  to  each,  but  not  like  placed,  are  symmetrical  triedral 
angles. 


172  GEOMETRY. 


BOOK  VII 


SOLID    GEOMETRY. 
DEFINITIONS. 

1.  A  Polyedron  is  a  solid,  or  volume,  bounded  on  all 
sides  by  planes.  The  bounding  planes  are  called  the 
faces  of  the  polyedron,  and  their  intersections  are  its 
edges. 

2.  A  Prism  is  a  polyedron,  having  two  of  its  faces, 
called  bases,  equal  polygons,  whose  planes  and  homolo- 
gous sides  are  parallel.  The  other,  or  lateral  faces,  are 
parallelograms,  and  constitute  the  convex  surface  of  the 
prism. 

The  bases  of  a  prism  are  distinguished  by  the  terms, 
upper  and  lower ;  and  the  altitude  of  the  prism  is  the  per- 
pendicular distance  between  its  bases. 

Prisms  are  denominated  triangular,  quadrangular,  pent- 
angular, etc.,  according  as  their  bases  are  triangles,  quad- 
rilaterals, pentagons,  etc. 

3.  A  Right  Prism  is  one  in  which  the  planes  of  the 
lateral  faces  are  perpendicular  to  the  planes  of  the  bases. 

4.  A  Parallelopipedon  is  a  prism 
whose  bases  are  parallelograms. 

5.  A  Rectangular  Parallelopipedon 
is  a  right  parallelopipedon,  with 
rectangular  bases. 


BOOK   VII 


178 


6.  A  Cube  or  Hexaedron  is  a  rectangu- 
lar parallelopipedon,  whose  faces  are  all 
equal  squares. 

7.  A  Diagonal  of  a  Polyedron  is  a  straight 
line  joining  the  vertices  of  two  solid 
angles  not  adjacent. 

8.  Similar  Polyedrons  are  those  which 

are  bounded  by  the  same  number  of  similar  polygons 
like  placed,  and  whose  solid  angles  are  equal  each  to 
each. 

Similar  parts,  whether  faces,  edges,  diagonals,  or 
angles,  similarly  placed  in  similar  polyedrons,  are  termed 
homologous. 

9.  A  Pyramid  is  a  polyedron,  having 
for  one  of  its  faces,  called  the  base,  any 
polygon  whatever,  and  for  its  other  faces 
triangles  having  a  common  vertex,  the 
sides  opposite  which,  in  the  several  trian- 
gles, being  the  sides  of  the  base  of  the 
pyramid. 

10.  The  Vertex  of  a  pyramid  is  the 
common  vertex  of  the  triangular  faces. 

11.  The  Altitude  of  a  pyramid  is  the  perpendicular 
distance  from  its  vertex  to  the  plane  of  its  base. 

12.  A  Right  Pyramid  is  one  whose  base  is  a  regular 
polygon,  and  whose  vertex  is  in  the  perpendicular  to  the 
base  at  its  center.  This  perpendicular  is  called  the  axis 
of  the  pyramid. 

13.  The  Slant  Height  of  a  right  pyramid  is  the  perpen- 
dicular distance  from  the  vertex  to  one  of  the  sides  of 
the  base. 

14.  The  Frustum  of  a  Pyramid  is  a  portion  of  the  pyr- 
amid included  between  its  base  and  a  section  made  by  a 
plane  parallel  to  the  base. 

Pyramids,  like  prisms,  are  named  from  the  forms  of 
their  bases. 
15* 


174 


GEOMETRY. 


15.  A  Cylinder  is  a  body,  having  for 
its  ends,  or  bases,  two  equal  circles, 
the  planes  of  which  are  perpendicular 
to  the  line  joining  their  centers ;  the 
remainder  of  its  surface  may  be  con- 
ceived as  formed  by  the  motion  of  a 
line,  which  constantly  touches  the  cir- 
cumferences of  the  bases,  while  it 
remains  parallel  to  the  line  which 
joins  their  centers. 

"We  may  otherwise  define  the  cylinder  as  a  body  gen- 
erated by  the  revolution  of  a  rectangle  about  one  of  its 
sides  as  an  immovable  axis. 

The  sides  of  the  rectangle  perpendicular  to  the  axis 
generate  the  bases  of  the  cylinder ;  and  the  side  opposite 
the  axis  generates  its  convex  surface.  The  line  joining 
the  centers  of  the  bases  of  the  cylinder  is  its  axis,  and  is 
also  its  altitude. 

If,  within  the  base  of  a  cylinder,  any  polygon  be  in- 
scribed, and  on  it,  as  a  base,  a  right  prism  be  con- 
structed, having  for  its  altitude  that  of  the  cylinder,  such 
prism  is  said  to  be  inscribed  in  the  cylinder,  and  the  cylin- 
der is  said  to  circumscribe  the  prism. 

Thus,  in  the  last  figure,  ABOBEc  is  an  inscribed 
prism,  and  it  is  plain  that  all  its  lateral  edges  are  con- 
tained in  the  convex  surface  of  the  cylinder 

If,  about  the  base  of  a  cylinder,  any 
polygon  be  circumscribed,  and  on  it, 
as  a  base,  a  right  prism  be  con- 
structed, having  for  its  altitude  that 
of  the  cylinder,  such  prism  is  said  to 
be  circumscribed  about  the  cylinder,  and 
the  cylinder  is  said  to  be  inscribed  in 
the  prism. 

Thus,  ABCBEFc  is  a  circum- 
scribed prism;   and  it  is  plain  that 


BOOK    VII 


175 


the  line,  w,  which  joins  the  points  of 
tangency  of  the  sides,  EF  and  ef,  with 
the  circumferences  of  the  bases  of  the 
cylinder,  is  common  to  the  convex  sur- 
faces of  the  cylinder  and  prism. 

16.  A  Cone  is  a  body  bounded  by  a 
circle  and  the  surface  generated  by  the 
motion  of  a  straight  line,  which  con- 
stantly passes  through  a  point  in  the 
perpendicular  to  the  plane  of  the  circle 
at  its  center,  and  the  different  points  in 
its  circumference. 

The  cone  may  be  otherwise  defined  as  a  body  gene- 
rated by  the  revolution  of  a  right-angled  triangle  about 
one  of  its  sides  as  an  immovable  axis.  The  other  side 
of  the  triangle  will  generate  the  base  of  the  cone,  while 
the  hypotenuse  generates  the  convex  surface. 

The  side  about  which  the  generating  triangle  revolves 
is  the  axis  of  the  cone,  and  is  at  the  same  time  its  altitude. 

If,  within  the  base  of  the  cone,  any 
polygon  be  inscribed,  and  on  it,  as  a 
base,  a  pyramid  be  constructed,  having 
for  its  vertex  that  of  the  cone,  such 
pyramid  is  said  to  be  inscribed  in  the 
cone,  and  the  cone  is  said  to  circumscribe 
the  pyramid. 

Thus,  in  the  accompanying  figure, 
V — ABODE,  is  an  inscribed  pyramid, 
and  it  is  plain  that  all  its  lateral  edges 
are  contained  in  the  convex  surface  of 
the  cone. 

If,  about  the  base  of  a  cone,  any  poly- 
gon be  circumscribed,  and  on  it,  as  a 
base,  a  pyramid  be  constructed,  having 
for  its  vertex  that  of  the  cone,  such  pyramid  is  said  to  be 
circumscribed  about  the  cone,  and  the  cone  is  said  to  be 
inscribed  in  the  pyramid. 


176 


GEOMETRY. 


17.  The  Frustum  of  a  Cone  is  the  portion  of  the  cone  that 
is  included  between  its  base  and  a  section  made  by  a  plane 
parallel  to  the  base. 

18.  Similar  Cylinders,  and  also  Similar  Cones,  are  such  as 
have  their  axes  proportional  to  the  radii  of  their  bases. 

19.  A  Sphere  is  a  body  bounded  by  one  uniformly-curved 
surface,  all  the  points  of  which  are  at  the  same  distance 
from  a  certain  point  within,  called  the  center. 

We  may  otherwise  define  the  sphere  as  a  body  gene- 
rated by  the  revolution  of  a  semicircle  about  its  diameter 
as  an  immovable  axis. 

20.  A  Spherical  Sector  is  that 
portion  of  a  sphere  which  is  in- 
cluded between  the  surfaces  of 
two  cones  having  their  verti- 
ces at  the  center  of  the  sphere. 
Or,  it  is  that  portion  of  the 
sphere  which  is  generated  by  a 
sector  of  the  generating  semi- 
circle. 

21.  The  Radius  of  a  Sphere  is 
a  straight  line  drawn  from  the 

center  to  any  point  in  the  surface ;  and  the  diameter  is 
a  straight  line  drawn  through  the  center,  and  limited  on 
both  sides  by  the  surface. 

All  the  diameters  of  a  sphere  are  equal,  each  being 
twice  the  radius. 

22.  A  Tangent  Plane  to  a  sphere  is  one  which  has  a 
single  point  in  the  surface  of  the  sphere,  all  the  others 
being  without  it. 

23.  A  Secant  Plane  to  a  sphere  is  one  which  has  more 
than  one  point  in  the  surface  of  the  sphere,  and  lies 
partly  within  and  partly  without  it. 

Assuming,  what  will  presently  be  proved,  that  the  in- 
tersection of  a  sphere  by  a  plane  is  a  circle, 

24.  A  Small  Circle  of  a  sphere  is  one  whose  plane  does 
not  pass  through  its  center;  and 


BOOK   VII. 


177 


25.  A  Great  Circle  of  a  sphere  is  one  whose  plane  passes 
through  the  center  of  the  sphere. 

26.  A  Zone  of  a  sphere  is  the  portion  of  its  surface  in- 
cluded between  the  circumferences  of  any  two  of  its  paral- 
lel circles,  called  the  bases  of  the  zone.  When  the  plane 
of  one  of  these  circles  becomes  tangent  to  the  sphere,  the 
zone  has  a  single  base. 

27.  A  Spherical  Segment  is  a  portion  of  the  volume  of  a 
sphere  included  between  any  two  of  its  parallel  circles, 
called  the  bases  of  the  segment. 

The  altitude  of  a  zone,  or  of  a  segment,  of  a  sphere, 
is  the  perpendicular  distance  between  the  planes  of  its 
bases. 

28.  The  area  of  a  surface  is  measured  by  the  product 
of  its  length  and  breadth,  and  these  dimensions  are  always 
conceived  to  be  exactly  at  right  angles  to  each  other. 

29.  In  a  similar  manner,  solids  are  measured  by  the 
product  of  their  length,  breadth,  and  height,  when  all  their 
dimensions  are  at  right  angles  to  each  other. 

The  product  of  the  length  and  breadth  of  a  solid,  is 
the  measure  of  the  surface  of  its  base. 

Let  P,  in  the  annexed  fig- 
ure, represent  the  measuring 
unit,  and  A F  the  rectangular 
solid'  to  be  measured. 

A  side  of  P  is  one  unit  in 
length,  one  in  breadth,  and 
one  in  height ;  one  inch,  one 
foot,  one  yard,  or  any  other  unit  that  may  be  taken. 


Then, 


lxlxl==l,  the  unit  cube. 


Now,  if  the  base  of  the  solid,  AC,  is,  as  here  repre- 
sented, 5  units  in  length  and  2  in  breadth,  it  is  obvious 
that  (5x2  =  10),  10  units,  each  equal  to  P,  can  be  placed 
on  the  base  of  AC,  and  no  more;  and  as  each  of  these 
units  will  occupy  a  unit  of  altitude,  therefore,  2  units  of 

M 


178 


GEOMETRY. 


altitude  will  contain  20  solid  units,  3  units  of  altitude, 
30  solid  units,  and  so  on ;  or,  in  general  terms,  the  num- 
ber of  square  units  in  the  base  multiplied  by  the  linear  units 
in  perpendicular  altitude,  will  give  the  solid  units  in  any  rect- 
angular solid. 


THEOREM    I. 

If  the  three  plane  faces  bounding  a  solid  angle  of  one  prism 
be  equal  to  the  three  plane  faces  bounding  a  solid  angle  of 
another,  each  to  each,  and  similarly  disposed,  the  prisms  will 
be  equal. 

Suppose  A  and  a  to  be  the  vertices  of  two  solid  angles, 
bounded  by  equal  and  similarly  placed  faces;  then  will 
the  prisms,  ABODE — iV'and  abcde — n,  be  equal. 

For,  if  we  place  the  base, 
abcde,  upon  its  equal,  the  base 
ABODE,  they  will  coincide; 
and  since  the  solid  angles, 
whose  vertices  are  A  and  a,  are 
equal,  the  lines  ab,  ae,  and  ap, 
respectively  coincide  with  AB, 
AE,  and  AP ;  but  the  faces,  al  and  ao,  of  the  one  prism, 
are  equal,  each  to  each,  to  the  faces,  AL  and  A  0,  of  the 
other;  therefore  pi  and  po  coincide  with  PL  and  PO, 
and  the  upper  bases  of  the  prisms  also  coincide :  hence, 
not  only  the  bases,  but  all  the  lateral  faces  of  the  two 
prisms  coincide,  and  the  prisms  are  equal. 

Oor.  If  the  two  prisms  are  right,  and  have  equal  bases 
and  altitudes,  they  are  equal.  For,  in  this  case,  the  rect- 
angular faces,  al  and  ao,  of  the  one,  are  respectively 
equal  to  the  rectangular  faces,  AL  and  AO,  of  the  other ; 
and  hence  the  three  faces  bounding  a  triedral  angle  in 
the  one,  are  equal  and  like  placed,  to  the  faces  bounding 
a  triedral  angle  in  the  other. 


BOOK    VII.  179 

THEOREM    II. 

The  opposite  faces  of  any  parallelopipedon  are  equal,  and 
their  planes  are  parallel. 

Let  ABOJ) — E  be  any  parallelopipedon ;  then  will  its 
opposite  faces  be  equal,  and  their  planes  will  be  parallel. 

The  bases  ABCB  and  FEGR  are 
equal,  and  their  planes  are  parallel, 
by  definitions  2  and  4  of  this  Book; 
it  remains  for  us,  therefore,  only  to 
show  that  any  two  of  the  opposite 
lateral  faces  are  equal  and  parallel. 

Since  all  the  faces  of  the  parallelopipedon  are  parallel- 
ograms, AB  is  equal  and  parallel  to  BO,  and  AH  is  also 
equal  and  parallel  to  BF;  hence  the  angles  HAB  and 
FBO  are  equal,  and  their  planes  are  parallel,  (Th.  17,  B. 
YI),  and  the  two  parallelograms,  HABGr  and  FBCE, 
having  two  adjacent  sides  and  the  included  angle  of  the 
one  equal  to  the  two  adjacent  sides  and  included  angle 
of  the  other,  are  equal. 

Cor.  1.  Hence,  of  the  six  faces  of  the  parallelopipedon, 
any  two  lying  opposite  may  be  taken  as  the  bases. 

Cor.  2.  The  four  diagonals  of  a  parallelopipedon  mutu- 
ally bisect  each  other.  For,  if  we  draw  AC  and  SB,  we 
shall  form  the  parallelogram  A  CEBl,  of  which  the  diago- 
nals are  AE  and  HC,  and  these  diagonals  are  at  the  same 
time  diagonals  of  the  parallelopipedon ;  but  the  diagonals 
of  a  parallelogram  mutually  bisect  each  other.  Now,  if 
the  diagonal  FB  be  drawn,  it  and  HC  will  bisect  each 
other,  since  they  are  diagonals  of  the  parallelogram 
FRBC.  In  like  manner  we  can  show  that  if  BG-  be 
drawn,  it  will  be  bisected  by  AE.  Hence,  the  four  diag- 
onals have  a  common  point  within  the  parallelopipedon. 

Scholium.  —  It  is  seen  at  once  that  the  six  faces  of  a  parallelopipe- 
don intersect  each  other  in  twelve  edges,  four  of  which  are  equal  to 
if  A,  four  to  AB,  and  four  to  AD.  Now,  we  may  conceive  the  parallel- 
opipedon to  be  bounded  by  the  planes  determined  by  the  three  linea 


180 


GEOMETRY. 


AH,  AB,  and  AD,  and  the  three  planes  passed  through  the  extremi- 
ties, H,  B,  and  D,  of  these  lines,  parallel  to  the  first  three  planes. 


THEOREM    III. 

The  convex  surface  of  a  right  prism  is  measured  by  the 
perimeter  of  its  base  multiplied  by  its  altitude. 

Let  ABODE — iVbe  a  right  prism,  of 
which  AP  is  the  altitude ;  then  will  its 
convex  surface  be  measured  by 

{AB  +  BO+CD  +  DE  +  JEA)  x  AP. 
For,  its  convex  surface  is  made  up  of  the 
rectangles  AL,  BM,  ON,  etc.,  and  each 
rectangle  is  measured  by  the  product  of 
its  base  by  its  altitude ;  but  the  altitude 
of  each  rectangle  is  equal  to  AP,  the  alti- 
tude of  the  prism ;  hence  the  convex  sur- 
face of  the  prism  is  measured  by  the  pro- 
duct of  the  sum  of  the  bases  of  the  rectangles,  or  the 
perimeter  of  the  base  of  the  prism,  by  the  common  alti- 
tude, AP. 

Cor.  Eight  prisms  will  have  equivalent  convex  surfaces, 
when  the  products  of  the  perimeters  of  their  bases  by 
their  altitudes  are  respectively  equal ;  and,  generally,  their 
convex  surfaces  will  be  to  each  other  as  the  products  of 
the  perimeters  of  their  bases  by  their  altitudes.  Hence, 
when  their  altitudes  are  equal,  their  surfaces  will  be  as 
the  perimeters  of  their  bases ;  and  when  the  perimeters 
of  their  bases  are  equal,  their  convex  surfaces  will  be  as 
their  altitudes. 


THEOREM   IV. 


The  two  sections  of  a  prism  made  by  parallel  planes  between 
its  bases  are  equal  polygons. 

Let  the  prism  ABODE  —  N  be  cut  between  its  bases 
by  two  parallel  planes,  making  the  sections  QBS,  etc., 


BOOK   VII 


181 


and  TVX,  etc. ;  then  will  these  sections 
be  equal  polygons. 

For,  since  the  secant  planes  are  paral- 
lel, their  intersections,  QR  and  TV,  by 
the  plane  of  the  face  UAPO  are  parallel, 
(Th.  10,  B.  VI) ;  and  being  included  be- 
tween the  parallel  lines,  AP  and  HO,  they 
are  also  equal.  In  the  same  manner  we 
may  prove  that  US  is  equal  and  parallel 
to  VX,  and  so  on  for  the  intersections  of 
the  secant  planes  by  the  other  faces  of 
the  prism.  Hence,  these  polygonal  sections  have  the 
sides  of  the  one  equal  to  the  sides  of  the  other,  each  to 
each.  The  angles  QEjS  and  TVX  are  equal,  because 
their  sides  are  parallel  and  lie  in  the  same  direction ;  and 
in  like  manner  we  prove  |_  RSY  =  [_  VXZ,  and  so  on 
for  the  other  corresponding  angles  of  the  polygons. 
Therefore,  these  polygons  are  both  mutually  equilateral 
and  mutually  equiangular,  and  consequently  are  equal. 

-  Cor.  A  section  of  a  prism  made  by  a  plane  parallel  to 
the  base  of  the  prism,  is  a  polygon  equal  to  the  base. 


THEOREM   V. 

Two  parallelopipedons,  the  one  rectangular  and  the  other 
oblique,  will  be  equal  in  volume  when,  having  the  same  base 
and  altitude,  two  opposite  lateral  faces  of  the  one  are  in  the 
planes  of  the  corresponding  lateral  faces  of  the  other. 

Designating  the  parallelo- 
pipedons  by  their  opposite 
diagonal  letters,  let  AGr  be 
the  rectangular,  and  AL  the 
oblique,  parallelopipedon,  hav- 
ing the  same  base,  AC,  and 
of  the  same  altitude,  namely, 

the  perpendicular  distance  be- 
16 


182  GEOMETRY. 

tween  the  parallel  planes,  A  0  and  EL.  Also  let  the  face, 
AK,  be  in  the  plane  of  the  face,  AF,  and  the  face,  BL,  in 
the  plane  of  the  face,  DGr.  We  are  now  to  prove  that  the 
oblique  parallelopipedon  is  equivalent  to  the  rectangular 
parallelopipedon. 

As  the  faces,  AF  and  AK,  are  in  the  same  plane,  and 
the  parallelopipedons  have  the  same  altitude,  FFK  is  a 
straight  line,  and  EF  —  IK,  because  each  is  equal  to  AB. 
If  from  the  whole  line,  EK,  we  take  EF,  and  then  from 
the  same  line  we  take  IK=  EF,  we  shall  have  the  re- 
mainders, Eland  FK,  equal ;  and  since  AE  and  BF  are 
parallel,  [_AEI  =  [_BFK;  hence  the  A's,  AEI  and 
BFK,  are  equal.  Since  HE  and  MI  are  both  parallel  to 
DA,  they  are  parallel  to  each  other,  and  EIMH  is  a  par- 
allelogram; for  like  reasons,  FKLGr  is  a  parallelogram, 
and  these  parallelograms  are  equal,  because  two  adjacent 
sides  and  the  included  angle  of  the  one  are  equal  to  two 
adjacent  sides  and  the  included  angle  of  the  other.  The 
parallelograms,  BE  and  OF,  being  the  opposite  faces  of 
the  parallelopipedon,  AGr,  are  equal.  Hence,  the  three 
plane  faces  bounding  the  triedral  angle,  E,  of  the  trian- 
gular prism,  EAI — H,  are  equal,  each  to  each,  and  like 
placed,  to  the  three  plane  faces  bounding  the  triedral, 
F,  of  the  triangular  prism,  FBK —  Cr,  and  these  prisms 
are  therefore  equal,  (Th.  1).  Now,  if  from  the  whole 
solid,  EABK — H,  we  take  the  prism,  EAI — H,  there 
will  remain  the  parallelopipedon,  AL;  and,  if  from  the 
same  solid,  we  take  the  prism,  FBK—Gr,  there  will  remain 
the  rectangular  parallelopipedon,  AG.  Therefore,  the 
oblique  and  the  rectangular  parallelopidon  are  equiva- 
lent. 

Cor.  The  volume  of  the  rectangular  parallelopipedon, 
AGr,  is  measured  by  the  base,  ABCB,  multiplied  by  the 
altitude,  AE,  (Def.  29) ;  consequently,  the  oblique  paral- 
lelopipedon is  measured  by  the  product  of  the  same  base 
by  the  same  altitude. 


BOOK    VII 


183 


Scholium. — If  neither  of  the  parallelopipedons  is  rectangular,  but 
they  still  have  the  same  base  and  the  same  altitude,  and  two  opposite 
lateral  faces  of  the  one  are  in  the  planes  of  the  corresponding  lateral 
faces  of  the  other,  by  precisely  the  same  reasoning  we  could  prove  the 
parallelopipedons  equivalent.  Hence,  in  general,  any  two  parallelo- 
pipedons will  be  equal  in  volume  when,  having  the  same  base  and  altitude, 
two  opposite  lateral  faces  of  the  one  are  in  the  planes  of  the  correspond- 
ing lateral  faces  of  the  other. 

THEOREM   VI. 

Two  parallelopipedons  having  equal  bases  and  equal  alti- 
tudes,  are  equivalent 

Let  AG  and  AL  be  two  paral- 
lelopipedons, having  a  common 
lower  base,  and  their  npper  bases 
in  the  same  plane,  HF.  Then 
will  these  parallelopipedons  be 
equivalent. 

Since  their  upper  bases  are  in 
the  same  plane,  the  lines  IM,  KL,  UF,  and  HG,  will 
intersect,  when  produced,  and  form  the  quadrilateral, 
NOPQ,  and  this  quadrilateral  will  be  a  parallelogram, 
(Cor.  2,  Th.  6,  B.  YI),  equal  to  the  common  lower  base 
of  the  two  parallelopipedons.  Now,  if  a  third  parallelo- 
pipedon  be  constructed,  having  BD  for  its  lower  base, 
and  OQ  for  its  upper  base,  it  will  be  equivalent  to  the  par- 
allelopipedon  AG-',  and  also  to  the  parallelopipedon  AL, 
(Th.  5,  Scholium) ;  hence,  the  two  given  parallelopipe- 
dons, being  each  equivalent  to  the  third  parallelopipe- 
don, are  equivalent  to  each  other. 

Hence,  two  parallelopipedons  having  equal  bases,  etc. 


THEOREM    VII. 

The  volume  of  any  parallelopipedon  is  measured  by  the 
product  of  its  base  and  altitude,  or  the  product  of  its  three 
dimensions. 


184 


GEOMETRY. 


h  H 

r 


9  G 


Let  ABCD — Q  be  any  parallelopipedon ;  then  will  its 
volume  be  expressed  by  the  product 
of  the  area  of  its  base  and  altitude. 

If  the  parallelopipedon  is  oblique, 
we  may  construct  on  its  base  a  right 
parallelopipedon,  by  erecting  perpen- 
diculars at  the  points  A,  B,  C,  and  D, 
and  making  them  each  equal  to  the 
altitude  of  the  given  parallelopipedon ; 
and  the  right  parallelopipedon,  thus 
constructed,  will  be  equivalent  to  the  given  parallelopip- 
edon, (Th.  6).  Now,  if  the  base,  ABCD,  is  a  rectangle, 
the  new  parallelopipedon  will  be  rectangular,  and  meas- 
ured by  the  product  of  its  base  and  altitude,  (Def.  16). 
But  if  the  base  is  not  rectangular,  let  fall  the  perpen- 
diculars, Be  and  Ad,  on  CD  and  CD  produced,  and  take 
the  rectangle  ABcd  for  the  base  of  a  rectangular  paral- 
lelopipedon, having  for  its  altitude  that  of  the  given 
parallelopipedon.  We  may  now  regard  the  rectangular 
face,  ABFU,  as  the  common  base  of  the  two  parallelo- 
pipedons,  Ag  and  AG-',  and,  as  they  have  a  common 
base,  and  equal  altitude,  they  are  equivalent.  Thus  we 
have  reduced  the  oblique  parallelopipedon,  first  to  an 
equivalent  right  parallelopipedon  on  the  same  base,  and 
then  the  right  to  an  equivalent  rectangular  parallelopip- 
edon on  an  equivalent  base,  all  having  the  same  alti- 
tude. But  the  rectangular  parallelopipedon,  Ag,  is 
measured  by  product  of  its  base,  ABcd,  and  its  altitude ; 
hence,  the  given  and  equivalent  oblique  parallelopipedon 
is  measured  by  the  product  of  its  equivalent  base  and 
equal  altitude. 

Hence,  the  volume  of  any  parallelopipedon,  etc. 

Cor.  Since  a  parallelopipedon  is  measured  by  the  pro- 
duct of  its  base  by  its  altitude,  it  follows  that  parallelo- 
pipedons  of  equivalent  bases,  and  equal  altitudes,  are  equiva- 
lent, or  equal  in  volume. 


BOOK   VII.  185 

THEOREM   VIII. 

Parallelopipedons  on  the  same,  or  equivalent  bases,  are  to 
each  other  as  their  altitudes ;  and  parallelopipedons  having 
equal  altitudes,  are  to  each  other  as  their  bases. 

Let  P  and  p  represent  two  parallelopipedons,  whose 
bases  are  denoted  by  B  and  b,  and  altitudes  by  A  and  a, 
respectively. 

Now,     P  =  B  x  A,  and  p  =  b  x  a,  (Th.  7). 

But  magnitudes  are  proportional  to  their  numerical 
measures ;  that  is, 

P  :  p  : :  B  x  A  :  b  X  a. 

If  the  bases  of  the  parallelopipedons  are  equivalent, 
we  have  B  =  b;  and  if  the  altitudes  are  equal,  we  have 
A  —  a.  Introducing  these  suppositions,  in  succession, 
in  the  above  proportion,  we  get 

P  :  p  : :  A  :  a, 
and  P  :  p  : :  B  :  b. 

Hence  the  theorem ;  Parallelopipedons  on  the  same,  etc. 

THEOREM    IX. 

Similar  parallelopipedons  are  to  each  other  as  the  cubes  of 
their  like  dimensions. 

Let  P  and  p  represent  any  two  similar  parallelopipe- 
dons, the  altitude  of  the  first  being  denoted  by  h,  and 
the  length  and  breadth  of  its  base  by  I  and  n,  respect- 
ively ;  and  let  h',  V,  and  nr,  in  order,  denote  the  corres- 
ponding dimensions  of  the  second. 
Then  we  are  to  prove  that 

P  :  p  ::  n*  :  nn  ::  P  :  J'3  ::  h*  :  h'\ 
We  have 

P  ==  Inh,  and  p  =  Vn'h'  (Th.  7) ; 
and  by  dividing  the   first  of  these   equations  by  the 
second,  member  by  member,  we  get 
16* 


186  GEOMETRY. 

P        Ink 


p       Vn'h' ' 
which,  reduced  to  a  proportion,  gives 
P  :  p  ::  Inh  :  Vn'V. 
But,  by  reason  of  the  similarity  of  the  parallelopipe- 
dons,  we  have  the  proportions 

I    :  V   : :  n  :  n' 
h  :  y  : :  n  :  n'; 
we  have  also  the  identical  proportion, 

n  :  n'  : :  n  :  n'. 
By  the  multiplication  of  these  proportions,  term  by 
term,  we  get,  (Th.  11,  B.  II), 

Inh  :  Vn'h'  : :  n*  :  n'3. 
That  is,  P  :  p  ::  n3  :  n'3. 

By  treating  in  the  same  manner  the  three  proportions, 
I  :  V  : :  h  :  h' 
n  :  nf  : :  h  :  hf 
h  :  h!  : :  h  :  h', 
we  should  obtain  the  proportion 

P  :  p  ::  h3  :  h'3', 
and,  by  a  like  process,  the  three  proportions, 
h  :  h!  :  :  I  :  V 
n  :  nf  : :  I  :  V 
X  V  V  : :  I  :  % 
will  give  us  the  proportion 

P  :  p  :  :  P  :  V3. 
Hence  the  theorem;  similar parallehpipedons  are  to  each 
other,  etc, 

THEOREM   X. 

The  two  triangular  prisms  into  which  any  parallelopipedon 
is  divided,  by  a  plane  passing  through  its  opposite  diagonal 
edges,  are  equivalent. 

Let  ABCD — F  be  a  parallelopipedon,  and  through 
the  diagonal  edges,  BF  and  DH,  pass  the  plane  BH,  divi- 
ding the  parallelopipedon  into  the  two  triangular  prisms, 


BOOK    VII.  187 

ABD — E  and  BOD — G- ;  then  we  are  to  prove  that  these 
prisms  are  equivalent.  Let  us  divide 
the  diagonal,  BD,  in  which  the  se- 
cant plane  intersects  the  base  of  the 
parallelopipedon,  into  three  equal 
parts,  a  and  c  being  the  points  of 
division.  In  the  base,  AB  CD,  con- 
struct the  complementary  paral- 
lelograms, a  0  and  a  A,  and  in  the 
parallelogram,  badD,  construct  the 
complementary  parallelograms, 
cd  and  cb,  and  conceive  these,  to- 
gether with  the  parallelograms, 
Ba,  ac,  cD,  to  be  the  bases  of 
smaller  parallelopipedons,  having 
their  lateral  faces  parallel  to  the 

lateral  faces  of,  and  their  altitude  equal  to  the  altitude  oi\ 
the  given  parallelopipedon,  AGr. 

Now  it  is  evident  that  the  triangular  prism,  BOD — Or, 
is  composed  of  the  parallelopipedons  on  the  bases,  aO 
and  cd,  and  the  triangular  prisms,  on  the  side  of  the 
secant  plane  with  this  prism,  into  which  this  plane  divides 
the  parallelopipedons  on  the  bases,  Ba,  ac,  and  cD.  The 
triangular  prism,  ABD — E,  is  also  composed  of  the  par- 
allelopipedons on  the  bases,  Aa  and  be,  together  with  the 
triangular  prisms  on  the  side  of  the  secant  plane  with 
this  prism,  into  which  this  plane  divides  the  parallelopip- 
edons on  the  bases,  Ba,  ac,  and  cD. 

But  the  parallelograms,  a  0  and  a  A,  being  complement- 
ary, are  equivalent,  (Th.  31,  B.  I) ;  and  for  the  same 
reason  the  parallelograms,  cd  and  cb,  are  equivalent ;  and 
since  parallelopipedons  on  equivalent  bases  and  of  equal 
altitudes,  are  equivalent,  (Cor.,  Th.  7),  we  have  the  sum 
of  parallelopipedons  on  bases  a 0  and  cd,  equivalent  to 
the  sum  of  parallelopipedons  on  the  bases,  aA  and  cb. 
Hence,  the  triangular  prisms,  ABD — E  and  BOD — #, 


188  GEOMETRY. 

differ  in  volume  only  by  the  difference  which  may  exist 
between  the  snms  of  the  triangular  prisms  on  the  two 
sides  of  the  secant  plane  into  which  this  plane  divides 
the  parallelopipedons  on  the  bases,  Ba,  ac,  and  cd. 

Now,  if  the  number  of  equal  parts  into  which  the  diag- 
onal is  divided,  be  indefinitely  multiplied,  it  still  holds 
true  that  the  triangular  prisms,  ABB — E  and  BOB — 6r, 
differ  in  volume  only  by  the  difference  between  the  sums 
of  the  triangular  prisms  on  the  two  sides  of  the.  secant 
plane  into  which  this  plane  divides  the  parallelopipedons 
constructed  on  the  bases  whose  diagonals  are  the  equal 
portions  of  the  diagonal,  BB.  But  in  this  case  the  sum 
of  these  parallelopipedons  themselves  becomes  an  indefi- 
nitely small  part  of  the  whole  parallelopipedon,  A  6r,  and 
the  difference  between  the  parts  of  an  indefinitely  small 
quantity  must  itself  be  indefinitely  small,  or  less  than 
any  assignable  quantity.  Therefore,  the  triangular 
prisms,  ABB — E  and  BOB — 6r,  differ  in  volume  by  less 
than  any  assignable  volume,  and  are  consequently  equiv- 
alent. 

Hence  the  theorem ;  the  two  triangular  prisms  into  which, 
etc. 

Cor.  1.  Any  triangular  prism,  as  ABB — E,  is  one  half 
the  parallelopipedon  having  the  same  triedral  angle,  A, 
and  the  same  edges,  AB,  AB,  and  AE. 

Cor.  2.  Since  the  volume  of  a  parallelopipedon  is  meas- 
ured by  the  product  of  its  base  and  altitude,  and  the  tri- 
angular prisms  into  which  it  is  divided  by  the  diagonal 
plane,  have  bases  equivalent  to  one  half  the  base  of  the 
parallelopipedon,  and  the  same  altitude,  it  follows  that, 
the  volume  of  a  triangular  prism  is  measured  by  the  product 
of  its  base  and  altitude. 

The  above  demonstration  is  less  direct,  but  is  thought 
to  be  more  simple,  than  that  generally  found  in  authors, 
and  which  is  here  given  as  a 


BOOK    VII. 


189 


Second  Demonstration. 

Let  ABQD — F  be  a  parallelo- 
pipedon,  divided  by  the  diagonal 
plane,  BH,  passing  through  the 
edges,  BF  and  Dff;  then  we  are 
to  prove  that  the  triangular 
prisms,  ABD—E  and  BCD— a, 
thus  formed,  are  equivalent. 

Through  the  points  B  and  F, 
pass  planes  perpendicular  to  the 
edge,  BF,  and  produce  the  late- 
ral faces  of  the  parallelopipedon 
to  intersect  the  plane  through  B ; 
then  the  sections  Bcda  and  Fghe 
are  equal  parallelograms.  For,  since  the  cutting  planes 
are  both  perpendicular  to  BF,  they  are  parallel,  (Th.  10, 
B.  VI) ;  and  because  the  opposite  faces  of  a  parallelo- 
pipedon are  in  parallel  planes,  (Th.  2),  and  the  intersec- 
tions of  two  parallel  planes  by  a  third  plane  are  parallel, 
(Th.  9,  B.  YI),  the  sections,  Bcda  and  Fghe,  are  equal 
parallelograms,  and  may  be  taken  as  the  bases  of  the 
right  parallelopipedon,  Bcda — h.  But  the  diagonal  plane 
divides  the  right  parallelopipedon  into  the  two  equal  tri- 
angular prisms,  aBd — e  and  Bed — g,  (Th.  1).  "We  will 
now  compare  the  right  prism  with  the  oblique  triangular 
prism  on  the  same  side  of  the  diagonal  plane. 

The  volume  ABB  —  e  is  common  to  the  two  prisms, 
ABB — E  and  aBd — e  ;  and  the  volume  eFh — E,  which, 
added  to  this  common  part,  forms  the  oblique  triangular 
prism,  is  equal  to  the  volume  aBd — A,  which,  added  to 
the  common  part,  forms  the  right  triangular  prism.  For, 
since  ABFE  and  aBFe  are  parallelograms,  AE  —  ae,  and 
taking  away  the  common  part  Ae,  we  have  aA=eE;  and 
since  BFHD  and  BFhd  are  parallelograms,  we  have  BH 
=  dh ;  and  from  these  equals  taking  away  the  common 
part  Dh,  we  have  dD  =  hH.    Now,  if  the  volume  eFh — B 


190  GEOMETRY. 

be  applied  to  the  volume  aBd —  D,  the  base  eFh  falling 
on  the  equal  base  aBd,  the  edges  eE  and  hH  will  fall 
upon  aA  and  dD  respectively,  because  they  are  perpen- 
dicular to  the  base  aBd,  (Cor.  2,  Th.  3,  B.  VI),  and  the 
point  E  will  fall  upon  the  point  A,  and  the  point  H  upon 
the  point  D ;  hence  the  volume  eFh — H  exactly  coincides 
with  the  volume  aBd — D,  and  the  oblique  triangular 
prism  ABB — E  is  equivalent  to  the  right  triangular 
prism  aBd — e. 

In  the  same  manner,  it  may  be  proved  that  the  oblique 
triangular  prism,  BCBG,i8  equivalent  to  the  right  tri- 
angular prism,  Bcdg.  The  oblique  triangular  prism  on 
either  side  of  the  diagonal  plane  is,  therefore,  equivalent 
to  the  corresponding  right  triangular  prism ;  and,  as  the 
two  right  triangular  prisms  are  equal,  the  oblique  trian- 
gular prisms  are  equivalent. 

Hence  the  theorem ;  the  two  triangular  prisms,  etc. 

THEOREM   XI. 

The  volume  of  any  prism  whatever  is  measured  by  the  prod- 
uct  of  the  area  of  its  base  and  altitude. 

For,  by  passing  planes  through  the  homologous  diag- 
onals of  the  upper  and  lower  bases  of  the  prism,  it  will 
be  divided  into  a  number  of  triangular  prisms,  each  of 
which  is  measured  by  the  product  of  the  area  of  its  base 
and  altitude.  Now,  as  these  triangular  prisms  all  have, 
for  their  common  altitude,  the  altitude  of  the  given 
prism,  when  we  add  the  measures  of  the  triangular 
prism,  to  get  that  of  the  whole  prism,  we  shall  have, 
for  this  measure,  the  common  altitude  multiplied  by  the 
sum  of  the  areas  of  the  bases  of  the  triangular  prisms : 
that  is,  the  product  of  the  area  of  the  polygonal  base 
and  the  altitude  of  the  prism. 

Hence  the  theorem ;  the  volume  of  any  prism,  etc. 

Cor.  If  A  denote  the  area  of  the  base,  and  H  the  alti- 


BOOK    VII.  191 

tude  of  a  prism,  its  volume  will  be  expressed  by  A  x  11. 
Calling  this  volume  F,  we  have 

V  4  A  x  H. 
Denoting  by  A',  W,  and  V,  in  order,  the  area  of  the 
base,  altitude,  and  volume  of  another  prism,  we  have 

V  =  A'  x  H'. 

Dividing  the  first  of  these  equations  by  the  second, 
member  by  member,  we  have 

V_         AxH 

V  '  A1  x  R1' 
which  gives  the  proportion, 

V  :    V  : :  A  x  E  :  Af  x  Hf. 
If  the  bases  are  equivalent,  this  proportion  becomes 

V  :  V  : :  H  :  Hr ; 
and  if  the  altitudes  are  equal,  it  reduces  to 

V  :    V  : :  A  :  A'. 
Hence,  prisms  of  equivalent  bases  are  to  each  other  as 
their  altitudes ;  and  prisms  of  equal  altitudes  are  to  each  other 
as  their  bases. 

THEOREM    XII. 

A  plane  passed  through  a  pyramid  parallel  to  its  base, 
divides  its  edges  and  altitude  proportionally,  and  makes  a 
section,  which  is  a  polygon  similar  to  the  base. 

Let  ABODE —  V  be  any  pyramid,  whose  base  is  in  the 
plane,  MN,  and  vertex  in  the  parallel  plane,  mn ;  and  let 
a  plane  be  passed  through  the  pyramid,  parallel  to  its 
base,  cutting  its  edges  at  the  points,  a,  5,  c,  d,  e,  and  the 
altitude,  JEF,  at  the  point  I.  By  joining  the  points,  a,  b, 
c,  etc.,  we  have  the  polygon  formed  by  the  intersection 
of  the  plane  and  the  sides  of  the  pyramid.  Now,  we  are 
to  prove  that  the  edges,  VA,  VB,  etc.,  and  the  altitude, 
FE,  are  divided  proportionally  at  the  points,  a,  b,  etc., 
and  Z;  and  that  the  polygon,  a,  b,  c,  d,  e,  is  similar  to  the 
base  of  the  pyramid. 


GEOMETRY. 


Since  the  cutting  plane  is  parallel  to  the  base  of  the 
pyramid,  ab  is  parallel  to  AB,  (Th.  9,  B.  VI) ;  for  the 
same  reason,  be  is  parallel  to  BO,  cd  to  OB,  etc.  Now, 
in  the  triangle  VAB,  because  ab  is  parallel  to  the  base 
AB,  we  have,  (Th.  17,  B.  II),  the  proportion, 

VA  :  Va  : :  VB  :  Vb. 

In  like  manner,  it  may  be  shown  that 

VB  :  Vb  : :  VO  :  Vc, 

and  so  on  for  the  other  lateral  edges  of  the  pyramid.  F 
being  the  point  in  which  the  perpendicular  from  E  pierces 
the  plane  mn,  and  I  the  point  in  which  the  parallel  secant 
plane  cuts  the  perpendicular,  if  we  join  the  points  F  and 
V,  and  also  the  points  I  and  e  by  straight  lines,  we  have 
in  the  triangle  FFV,  the  line  le  parallel  to  the  base  FV; 
hence  the  proportion 

VF  :  Ve  : :  FE  :  Fl. 
Therefore,  the  plane  passed  through  the  pyramid  par- 
allel to  its  base,  divides  the  altitude  into  parts  \diich  have 


BOOK    VII.  193 

to  each  other  the  same  ratio  as  the  parts  into  which  it 
divides  the  edges. 

Again,  since  ab  is  parallel  to  AB,  and  be  to  BO,  the 
angle  abc  is  equal  to  the  angle  ABO,  (Th.  8,  B.  I);  in 
the  same  manner  we  may  show  that  each  angle  in  the 
polygon,  abode,  is  equal  to  the  corresponding  angle  in  the 
polygon,  ABODE;  therefore  these  polygons  are  mutually 
equiangular.  But,  because  the  triangles  VBA  and  Vba 
are  similar,  their  homologous  sides  give  the  proportion 

Vb  :  VB  ::  ab  :  AB; 
and  because  the  triangles  Vbc  and  VBO  are  similar,  we 
also  have  the  proportion 

Vb  :  VB  : :  be  :  BO. 

Since  the  first  couplet  in  these  two  proportions  is  the 
same,  the  second  couplets  are  proportional,  and  give 
ab  :  AB  : :  be  :  BO. 

By  a  like  process,  we  can  prove  that 
be  :  BO  : :  cd  :   OB, 

and  that  cd  :   OD  ::  de  :  BE, 

and  so  on,  for  the  other  homologous  sides  of  the  two 
polygons. 

Hence,  the  two  polygons  are  not  only  mutually  equi- 
angular, but  the  sides  about  the  equal  angles  taken  in  the 
same  order  are  proportional,  and  the  polygons  are  there- 
fore similar,  (Def.  16,  B.  II). 

Hence  the  theorem;  a  plane  passed  through  a  pyramid, 
etc. 

Oor.  1.  Since  the  areas  of  similar  polygons  are  to  each 
other  as  the  squares  of  their  homologous  sides,  (Th.  22, 
B.  II),  we  have 

area  abode  :  area  ABODE  :  ab2  :  AB*. 

But,  ab  :  AB  ::  Va  :  VA    ::  Fl   :  FE; 

hence,  ab2  :  AB2  i:~Ff  :  FE2 : 

therefore,      area  abode  :  area  ABODE  :  Fl2  :  FE  . 
17  N 


194  GEOMETRY. 

That  is,  the  area  of  the  section  made  by  a  plane  passing 
through  a  pyramid  parallel  to  its  base,  is  to  the  area  of  the 
base,  as  the  perpendicular  distance  from  the  vertex  of  the 
pyramid  to  the  section,  is  to  the  altitude  of  the  pyramid. 

Cor.  2.  Let  V— ABODE  and  X—RST  be  two  pyra- 
mids, having  their  bases  in  the  plane  MN,  and  their  ver- 
tices in  the  parallel  plane  mn ;  and  suppose  a  plane  to  be 
passed  through  the  two  pyramids  parallel  to  the  common 
plane  of  their  bases,  making  in  the  one  the  section  abcde, 
and  in  the  other  the  section  rst. 


Now,  arenABCDE:  area  abcde  ::AB  :  ab ,  (Th.22,B.II), 
and      "  RST:    "        rst  ::RS2  :rs\ 

But,  AB  :  ab  : :    VB  :  Vb, 

and  RS  :  rs  : :  XR  :  Xr. 

Because  the  plane  which  makes  the  sections  is  parallel 
to  the  planes  MN  and  mn,  we  have,  (Th.  11,  B.  VI), 
VB  :  Vb  ::  XR  :  Xr; 
therefore,  (Cor.  2,  Th.  6,  B.II),  AB  i  ab  : :  RS  :  rs. 


:2  ~T2       ~~ 777*2 


By  squaring,        AB    :  ab    :  RS    :  rs  ; 

hence,    area  ABCDE :  area  abcde  : :  area  RST :  area  rst. 

That  is,  if  two  pyramids  having  equal  altitudes,  and  their 
bases  in  the  same  plane,  be  cut  by  a  plane  parallel  to  the  com- 
mon plane  of  their  bases,  the  areas  of  the  sections  will  be 
proportional  to  the  areas  of  the  bases ;  and  if  the  bases  are 
equivalent,  the  sections  will  also  be  equivalent. 

THEOREM    XIII. 

If  two  triangular  pyramids  have  equivalent  bases  and 
equal  altitudes,  they  are  equal  in  volume. 

Let  V — ABC  and  v — abc  be  two  triangular  pyramids, 
having  the  equivalent  bases,  ABC  and  abc,  and  let  the 
altitude  of  each  be  equal  to  CX;  then  will  these  two 
pyramids  be  equivalent. 


BOOK   VII 


195 


-  fI 

//  V     ' 

/It   \l 

1      JT^     1  '''  ' 

1    1    r\    1 

1  //    V 

/   h/\  i 

1         /  7            \  / 

xl       1/          VH 

A    //   S\  let 

Place  the  bases  of  the  pyramids  on  the  same  plane, 
with  their  vertices  in  the  same  direction,  and  divide  the 
altitude  into  any  number  of  equal  parts.  Through  the 
points  of  division  pass  planes  parallel  to  the  plane  of  the 
bases ;  the  corresponding  sections  made  in  the  pyramids 
by  these  planes  are  equivalent,  (Th.  12,  Cor.  2) ;  that  is, 
the  triangle  DEI?  is  equivalent  to  the  triangle  def,  the 
triangle  GrHI  to  the  triangle  ghi,  etc. 

Now,  let  triangular  prisms  be  constructed  on  the  tri- 
angles ABO,  DEF,  etc.,  of  the  pyramid  V— ABO,  these 
prisms  having  their  lateral  edges  parallel  to  the  edge, 
VO,  of  the  pyramid,  and  the  equal  parts  of  the  altitude, 
OX,  for  their  altitudes.  Portions  of  these  prisms  will  be 
exterior  to  the  pyramid  V — ABO,  and  the  sum  of  their 
volumes  will  exceed  the  volume  of  the  pyramid. 

On  the  bases  def,  ghi,  etc.,  in  the  other  pyramid,  con- 
struct interior  prisms, ,  as  represented  in  the  figure, 
their  lateral  edges  being  parallel  to  vc,  and  their  alti- 
tudes also  the  equal  parts  of  the  altitude,  OX.  Portions 
of  the  pyramid,  v — abe,  will  be  exterior  to  these  prisms, 


196  GEOMETRY. 

and  the  volume  of  the  pyramid  will  exceed  the  sum  of 
the  volumes  of  the  prisms. 

Since  the  sum  of  the  exterior  prisms,  constructed  in 
connection  with  the  pyramid  V — ABO,  is  greater  than 
the  pyramid,  and  the  sum  of  the  interior  prisms,  con- 
structed in  connection  with  the  pyramid  v — abc,  is  less 
than  this  pyramid,  it  follows  that  the  difference  of  these 
sums  is  greater  than  the  difference  of  the  pyramids  them- 
selves. But  the  second  exterior  prism,  or  that  on  the 
base  DEF,  is  equivalent  to  the  first  interior  prism,  or 
that  on  the  base  def,  and  the  third  exterior  prism  is 
equivalent  to  the  second  interior  prism,  (Th.  10,  Cor.  2), 
and  so  on.  That  is,  beginning  with  the  second  prism  from 
the  base  of  the  pyramid,  V — ABO,  and  taking  these 
prisms  in  order  towards  the  vertex  of  the  pyramid,  and 
comparing  them  with  the  prisms  in  the  pyramid,  v — abc, 
beginning  with  the  lowest,  and  taking  them  in  order 
toward  the  vertex  of  this  pyramid,  we  find  that  to  each 
exterior  prism  of  the  pyramid,  V — ABO,  exclusive  of 
the  first  or  lowest,  there  is  a  corresponding  equivalent 
interior  prism  in  the  pyramid,  v — abc. 

Hence  the  prism,  ABODEF,  is  the  difference  between 
the  sum  of  the  prisms  constructed  in  connection  with 
the  pyramid,  V— ABO,  and  the  sum  of  the  interior 
prisms  constructed  in  the  pyramid,  v — abc.  But  the  first 
sum  being  a  volume  greater  than  the  pyramid,  V — ABO, 
and  the  second  sum  a  volume  less  than  the  pyramid, 
v — abc,  it  follows  that  the  volumes  of  the  pyramids  differ 
by  less  than  the  prism,  ABODEF. 

Now,  however  great  the  number  of  equal  parts  into 
which  the  altitude,  OX,  be  divided,  and  the  correspond- 
ing number  of  prisms  constructed  in  connection  with 
each  pyramid,  it  would  still  be  true  that  the  difference 
between  the  volumes  of  the  pyramids  would  be  less  than 
the  volume  of  the  lowest  prism  of  the  pyramid  V—  ABO', 
but  when  we  make  the  number  of  equal  parts  into  which 


BOOK    VII.  .  197 

the  altitude  is  divided  indefinitely  great,  the  volume  of 
this  prism  becomes  indefinitely  small :  that  is,  the  differ- 
ence between  the  volumes  of  the  pyramids  is  less  than 
an  indefinitely  small  volume ;  or,  in  other  words,  there 
is  no  assignable  difference  between  the  two  pyramids, 
and  they  are,  therefore,  equivalent. 
Hence  the  theorem ;  if  two  triangular  pyramids,  etc. 

THEOREM    XIV. 

Any  triangular  pyramid  is  one  third  of  the  triangular 
prism  having  the  same  base  and  equal  altitude. 

Let  F — ABC  be  a  triangular  pyramid,  and  through  F 
pass  a  plane  parallel  to  the  plane  of  the  base,  ABC.    In 
this  plane,  through  F,  construct  the 
triangle,  FDE,  having  its  sides,  FD,  E 

DF,  and  FF,  parallel  and  equal  to  B  C,  7vnT     ~y7\ 

CA,  and  AB,  respectively.     The  tri-  /     \/*\      / 

angle,  FDF,  may  be  taken  as  the  /  /\  \  \ 
upper  base  of  a  triangular  prism  of  \/  /  \\/ 
which  the  lower  base  is  ABC.  ^\  ~\  / 

IsTow,  this  triangular  prism  is  com-  13 

posed  of  the  given  triangular  pyramid, 
F — ABC,  and  of  the  quadrangular  pyramid,  F — A  CDF. 
This  last  pyramid  may  be  divided  by  a  plane  through  the 
three  points,  C,  F,  and  F,  into  the  two  triangular  pyra- 
mids, F—DFC  and  F—ACF.  But  the  pyramid,  jP— 
BFC,  may  be  regarded  as  having  the  triangle,  FFB, 
equal  to  the  triangle,  ABC,  for  its  base,  and  the  point,  C, 
for  its  vertex.  The  two  pyramids,  F—AB  C  and  C—JDFF, 
have  equal  bases  and  equal  altitudes ;  they  are  therefore 
equivalent,  (Th.  13).  Again,  the  two  pyramids,  F—DFC 
and  F — ACE,  have  a  common  vertex,  and  equivalent  bases 
in  the  same  plane,  and  they  are  also  equivalent.  There- 
fore, the  triangular  prism,  ABCDEF,  is  composed  of 
17* 


198 


GEOMETRY. 


three  equivalent  triangular  pyramids,  one  of  which  is  the 
given  triangular  pyramid,  F — ABC. 

Hence  the  theorem;  any  triangular  pyramid  is  one  third 
of  the  triangular  prism,  etc. 

Cor.  The  volume  of  the  triangular  prism  being  meas- 
ured by  the  product  of  its  base  and  altitude,  the  volume  of 
a  triangular  pyramid  is  measured  by  one  third  of  the  product 
of  its  base  and  altitude. 

THEOREM    XV, 

The  volume  of  any  pyramid  whatever  is  measured  by  one 
third  of  the  product  of  its  base  and  altitude. 

Let  V — ABCDU  be  any  pyramid ;  then  will  its  volume 
be  measured  by  one  third  of  the  product  of  its  base  and 
altitude. 

In  the  base  of  the  pyramid,  draw  the 
diagonals,  AB  and  AC,  and  through 
its  vertex  and  these  diagonals,  pass 
planes,  thus  dividing  the  pyramid  into 
a  number  of  triangular  pyramids 
having  the  common  vertex  V,  and  the 
altitude  of  the  given  pyramid  for  their 
common  altitude. 

Now,  each  of  these  triangular  pyra- 
mids is  measured  by  one  third  of 
the  product  of  its  base  and  altitude, 
(Cor.,  Th.  14),  and  their  sum,  which 
constitutes  the  polygonal  pyramid,  is 
therefore  measured  by  one  third  "of 
the  product  of  the  sum  of  the  trian- 
gular bases  and  the  common  altitude ;  but  the  sum  of  the 
triangular  bases  constitutes  the  polygonal  base,  ABCDE. 

Hence  the  theorem ;  the  volume  of  any  pyramid  what- 
ever, etc. 

Cor.  1.  Denote,  by  B,  H,  and  V,  respectively,  the  base, 
altitude,  and  volume  of  one  pyramid,  and  by  B',  W,  and 


BOOK   VII.  199 

F7,  the  base,  altitude,  and  volume  of  another ;  then  we 
shall  have 

V  =  $B  x  IT, 
and                         V  =  \B'  x  W. 

Dividing  the  first  of  these  equations  by  the  second, 
member  by  member,  we  have 

V  -B  x  H 
Vf     B'  x  W' 

which,  in  the  form  of  a  proportion,  gives 
V  :    V  : :  B  X  H  :  B'  X  W. 

From  this  proportion  we  deduce  the  following  conse- 
quences : 

1st.  Pyramids  are  to  each  other  as  the  products  of  their 
bases  and  altitudes. 

2d.  Pyramids  having  equivalent  bases  are  to  each  other  as 
their  altitudes. 

3d.  Pyramids  having  equal  altitudes  are  to  each  other  as 
their  bases. 

Cor.  2.  Since  a  prism  is  measured  by  the  product  of 
it's  base  and  altitude,  and  a  pyramid  by  one  third  of  the 
product  of  its  base  and  altitude,  we  conclude  that  any 
pyramid  is  one  third  of  a  prism  having  an  equivalent  base  and 
equal  altitude. 

THEOREM    XVI. 

The  volume  of  the  frustum  of  a  pyramid  is  equivalent  to 
the  sum  of  the  volumes  of  three  pyramids,  each  of  which  has 
an  altitude  equal  to  that  of  the  frustum,  and  whose  bases  are, 
respectively,  the  lower  base  of  the  frustum,  the  upper  base  of 
the  frustum,  and  a  mean  proportional  between  these  bases. 

Let  V—  ABODE  and  X—RST  be  two  pyramids,  the 
one  polygonal  and  the  other  triangular,  having  equiva- 
lent bases  and  equal  altitudes ;  and  let  their  bases  be 
placed  on  the  plane  MN,  their  vertices  falling  on  the 
parallel  plane  mn.     Pass  through  the  pyramids  a  plane 


GEOMETRY. 


parallel  to  the  common  plane  of  their  bases,  cutting  out 
the  sections  abode  and  rst ;  these  sections  are  equivalent, 
(Th.  12,  Cor.  2),  and  the  pyramids,  V — abode  and  X — rst, 
are  equivalent,  (Th.  13).  Now,  since  the  pyramids, 
V— ABODE  and  X—EST,  are  equivalent,  if  from  the 
first  we  take  the  pyramid,  V — abode,  and  from  the  second, 
the  pyramid,  X — rst,  the  remainders,  or  the  frusta, 
ABODE — a  and  BST—r,  will  be  equivalent. 

If,  then,  we  prove  the  theorem  in  the  case  of  the  frus- 
tum of  a  triangular  pyramid,  it  will  be  proved  for  the 
frustum  of  any  pyramid  whatever. 

Let  ABO—D  be  the  frustum  of  a 
triangular  pyramid.  Through  the 
points  D,  B,  and  0,  pass  a  plane, 
and  through  the  points  D,  C,  and 
E,  pass  another,  thus  dividing  the 
frustum  into  three  triangular  pyra- 
mids, viz.,  D—ABO,  O—DEF,  and 
D—BEO. 

Now,  the  first  of  these  has,  for  its 


BOOK   VII.  201 

base,  the  lower  base  of  the  frustum,  and  for  its  altitude 
the  altitude  of  the  frustum,  since  its  vertex  is  in  the 
upper  base ;  the  second  has,  for  its  base,  the  upper  base 
of  the  frustum,  and  for  its  altitude  the  altitude  of  the 
frustum,  since  its  vertex  is  in  the  lower  base.  Hence, 
these  are  two  of  the  three  pyramids  required  by  the 
enunciation  of  the  theorem ;  and  we  have  now  only  to 
prove  that  the  third  is  equivalent  to  one  having,  for  its 
base,  a  mean  proportional  between  the  bases  of  the  frus- 
tum, and  an  altitude  equal  to  that  of  the  frustum. 

In  the  face  ABED,  draw  HB  parallel  to  BE,  and 
draw  HE  and  HO.  The  two  pyramids,  B — BEO  and 
H—BEO,  are  equivalent,  since  they  have  a  common 
base  and  equal  altitudes,  their  vertices  being  in  the  line 
BH,  which  is  parallel  to  the  plane  of  their  common 
base,  (Th.  7,  B.  VI).  We  may,  therefore,  substitute  the 
pyramid,  H—BEO,  for  the  pyramid,  D—BEC.  But  the 
triangle,  BOH,  may  be  taken  as  the  base,  and  E  as  the 
vertex  of  this  new  pyramid ;  hence,  it  has  the  required 
altitude,  and  we  must  now  prove  that  it  has  the  required 
base. 

The  triangles,  ABO  and  HBO,  have  a  common  vertex, 
and  their  bases  in  the  same  line ;  hence,  (Th.  16,  B.  II), 

A  ABO  :  A  HBO  : :  AB  :  HB  ::  AB  :  BE.    (1) 

In  the  triangles,  BEE  and  HBO,  [__  E  =  L  -#,  and 
BE=HB',  hence,  if  BEE  be  applied  to  HBO,  [__  E  fil- 
ing on  [_  B,  and  the  side  BE  on  HB,  the  point  B  will 
fall  on  H,  and  the  triangles,  in  this  position,  will  have  a 
common  vertex,  H,  and  their  bases  in  the  same  line ; 
hence 

A  HBO  :  A  BEE  : :  BO  :  EF.     (2) 

But,  because  the  triangles,  ABO  and  BEE,  are  similar, 
we  have 

AB  :  BE  ::  BO  :  EF.     (3) 

From  proportions  (1),  (2)?  and  (3),  we  have,  (Th.  6, 
B.  II), 


202  GEOMETRY. 

A  ABO  :  A  HBO  : :  A  HBO  :  A  DBF; 
that  is,  the  base,  HBO,  is  a  mean  proportional  between 
the  lower  and  upper  bases  of  the  frustum. 

Hence  the  theorem ;  the  volume  of  the  frustum  of  a  pyra- 
mid, etc. 

THEOREM    XVII. 

The  convex  surface  of  any  right  pyramid  is  measured  by 
the  perimeter  of  its  base,  multiplied  by  one  half  its  slant  height. 

Let  S— ABODE F  be  a  right  pyramid, 
of  which  SH  is  the  slant  height ;  then  will 
its  convex  surface  have,  for  its  measure, 

±SH{AB  +  BO+  OD  +  DB+EF+  FA). 

Since  the  base  is  a  regular  polygon,  and 
the  perpendicular,  drawn  to  its  plane  from 
S,  passes  through  its  center,  the   edges, 
SA,  SB,  SO,  etc.,  are  equal,  (Cor.  Th.  4,        ah~b 
B.  VI),  and  the  triangles  SAB,  SBO,  etc.,  are  equal,  and 
isosceles,  each  having  an  altitude  equal  to  SH. 

Now,  AB  x  %SH  measures  the  area  of  the  triangle, 
SAB ;  and  BO  x  %SH  measures  the  area  of  the  triangle, 
SBO;  and  so  on,  for  the  other  triangular  faces  of  the 
pyramid.  By  the  addition  of  these  different  measures, 
we  get 

iSH(AB  +  BO+OD  +  DH+UF+  FA), 

as  the  measure  of  the  total  convex  surface  of  the  pyramid. 
Hence  the  theorem;   the  convex  surface  of  any  right 
pyramid,  etc. 

THEOREM  XVIII. 

The  convex  surface  of  the  frustum  of  any  right  pyramid  is 
measured  by  the  sum  of  the  perimeters  of  the  two  bases,  mul- 
tiplied by  one  half  the  slant  height  of  the  frustum. 

Let  ABODEF — d  be  the  frustum  of  a  right  pyramid ; 
then  will  its  convex  surface  be  measured  by 

iHh{AB+BC+CD+DI!+EF+FA+ab+b<>±cd+de+tf+fa). 


BOOK    VII. 


203 


e 

/    Xffl  i 

d 
hbV 

W  1 

|] 

1 
l 

V 

For,  the  upper  base,  abcdef,  of  the 
frustum  is  a  section  of  a  pyramid 
by  a  plane  parallel  to  the  lower 
base,  (Def.  14),  and  is,  therefore, 
similar  to  the  lower  base,  (Th.  12). 
But  the  lower  base  is  a  regular 
polygon,  (Def.  12);  hence,  the  up- 
per base  is  also  a  regular  polygon, 
of  the  same  name;  and  as  ab  and 
AB  are  intersections  of  a  face  of 
the  pyramid  by  two  parallel  planes,  A     ST  B 

they  are  parallel.  For  the  same  reason,  be  is  parallel  to 
BO,  cd  to  OB,  etc.,  and  the  lateral  faces  of  the  frustum 
are  all  equal  trapezoids,  each  having  an  altitude  equal 
to  ITJi,  the  slant  height  of  the  frustum. 

The  trapezoid  ABba  has,  for  its  measure,  %Hh(AB+ab), 
(Th.  34,  Book  I) ;  the  trapezoid  BCcb  has,  for  its  meas- 
ure, %Hh(BC  +  be),  and  so  on,  for  the  other  lateral  faces 
of  the  frustum. 

Adding  all  these  measures,  we  find,  for  their  sum, 
which  is  the  whole  convex  surface  of  the  frustum, 

\Rh  {AB+BC+  CD+DE+EF+  FA+db+bc+cd+de+ef+fa). 

Hence  the  theorem ;  the  convex  surface  of  the  frustum, 
etc. 


THEOREM    XIX. 

The  volumes  of  similar  triangular  prisms  are  to  each  other 
as  the  cubes  constructed  on  their  homologous  edges. 

Let  ABC— I7  and 
abp— /be two  similar 
triangular  prisms ; 
then  will  their  vol- 
umes be  to  each 
other  as  the  cubes, 
whose  edges  are  the 
homologous    edges 


204  GEOMETRY. 

AB  and  ab,  or  as  the  cubes,  whose  edges  are  the  homol- 
ogous edges  BE  and  be,  etc.  Since  the  prisms  are  similar, 
the  solid  angles,  whose  vertices  are  B  and  b,  are  equal; 
and  the  smaller  prism,  when  so  applied  to  the  larger  that 
these  solid  angles  coincide,  will  take,  within  the  larger, 
the  position  represented  by  the  dotted  lines,  In  this 
position  of  the  prisms,  draw  EH  perpendicular  to  the 
plane  of  the  base  ABO,  and  join  the  foot  of  the  perpen- 
dicular to  the  point  B,  and  in  the  triangle  BEH  draw, 
through  e,  the  line  eh,  parallel  to  EH;  then  will  EH 
represent  the  altitude  of  the  larger  prism,  and  eh  that  of 
the  smaller. 

J^ow,  as  the  bases  ABC  and  aBc,  are  homologous  faces, 
they  are  similar,  and  we  have,  (Th.  20,  Book  II), 

A  ABC  :  A  aBc  ::  AB*  :~a~B2  (1) 

But  the  A's  BEH  and  Beh  are  equiangular,  and  there- 
fore similar,  and  their  homologous  sides  give  the  propor- 
tion 

BE  :  Be  ::  EH  :  eh  (2) 

and  from  the  homologous   sides   of  the  similar  faces, 
ABED  and  aBed,  we  also  have 

BE  :  Be  ::  AB  :  aB  (3) 

Proportions  (2)  and  (3 ),  having  an  antecedent  and  con- 
sequent the  same  in  both,  we  have,  (Th.  6,  B.  II), 

EH  :  eh  ::  AB  :  aB  (4) 

By  the  multiplication  of  proportions  (1)  and  (4)?  term 
by  term,  we  get 

A  ABC  X  EH:  A  aBc  X  eh::  AB3  :  aB3 

But  A  ABC  x  EH  measures  the  volume  of  the  larger 
prism,  and  A  aBc  x  eh  measures  the  volume  of  the 
smaller. 

Hence  the  theorem;  the  volumes  of  similar  triangular 
prisms,  etc. 


BOOK    VII.  205 

Cor.  1.  The  volumes  of  two  similar  prisms  having  any 
bases  whatever,  are  to  each  other  as  the  cubes  constructed  on 
their  homologous  edges. 

For,  if  planes  be  passed  through  any  one  of  the  lateral 
edges,  and  the  several  diagonal  edges,  of  ,one  of  these 
prisms,  this  prism  will  be  divided  into  a  number  of  smaller 
triangular  prisms.  Taking  the  homologous  edge  of  the 
other  prism,  and  passing  planes  through  it  and  the  seve- 
ral diagonal  edges,  this  prism  will  also  be  divided  into 
the  same  number  of  smaller  triangular  prisms,  similar  to 
those  of  the  first,  each  to  each,  and  similarly  placed. 

Kow,  the  similar  smaller  prisms,  being  triangular,  are 
to  each  other  as  the  cubes  of  their  homologous  edges ; 
and  being  like  parts  of  the  larger  prisms,  it  follows  that 
the  larger  prisms  are  to  each  other  as  the  cubes  of  the 
homologous  edges  of  any  two  similar  smaller  prisms.  But 
the  homologous  edges  of  the  similar  smaller  prisms  are 
to  each  other  as  the  homologous  edges  of  tlie  given 
prisms ;  hence  we  conclude  that  the  given  prisms  are  to 
each  other  as  the  cubes  of  their  homologous  edges. 

Cor.  2.  The  volumes  of  two  similar  pyramids  having  any 
bases  whatever,  are  to  each  other  as  the  cubes  constructed  on 
their  homologous  edges. 

For,  since  the  pyramids  are  similar,  their  bases  are 
similar  polygons ;  and  upon  them,  as  bases,  two  similar 
prisms  may  be  constructed,  having  for  their  altitudes,  the 
altitudes  of  their  respective  pyramids,  and  their  lateral 
edges  parallel  to  any  two  homologous  lateral  edges  of  the 
pyramids. 

Now,  these  similar  prisms  are  to  each  other  as  the  cubes 
of  their  homologous  edges,  which  may  be  taken  as  the 
homologous  sides  of  their  bases,  or  as  their  lateral  edges, 
which  were  taken  equal  and  parallel  to  any  two  arbi- 
trarily assumed  homologous  lateral  edges  of  the  two 
pyramids ;  hence  the  pyramids  are  to  each  other  as  the 
cubes  constructed  on  any  two  homologous  edges. 
18 


206 


GEOMETRY. 


Cor.  3.  The  volumes  of  any  two  similar  polyedrons  are  to 
each  other  as  the  cubes  constructed  on  their  homologous  edges. 

For,  by  passing  planes  through  the  vertices  of  the 
homologous  solid  angles  of  such  polyedrons,  they  may 
both  be  divided  into  the  same  number  of  triangular 
pyramids,  those  of  the  one  similar  to  those  of  the  other, 
each  to  each,  and  similarly  placed. 

Now,  any  two  of  these  similar  triangular  pyramids  are 
to  each  other  as  the  cubes  of  their  homologous  edges ; 
and  being  like  parts  of  their  respective  polyedrons,  it 
follows  that  the  polyedrons  are  to  each  other  as  the  cubes 
of  the  homologous  edges  of  any  two  of  the  similar  tri- 
angular pyramids  into  which  they  may  be  divided.  But 
the  homologous  edges  of  the  similar  triangular  pyramids 
are  to  each  other  as  the  homologous  edges  of  the  poly- 
edrons ;  hence  the  polyedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 


THEOREM    XX. 


!!J~£X 


The  convex  surface  of  the  frustum  of  a  cone  is  measured 
by  the  product  of  the  slant  height  and  one  half  the  sum  of 
the  circumferences  of  the  bases  of  the  frustum. 

Let  ABOB — abed  be  the  frustum  of 

a  cone ;  then  will  its  convex  surface  be 

t  ,      A         (circ.  00  4-  circ.  oc) 
measured  by  Aa  x ~ '-, 

in  which  the  expression,  circ.  00,  de- 
notes the  circumference  of  the  circle 
of  which  00  is  the  radius.  Inscribe  in 
the  lower  base  of  the  frustum,  a  regu- 
lar polygon  haviug  any  number  of 
sides,  and  in  the  upper  base  a  similar 
polygon,  having  its  sides  parallel  to 
those  of  the  polygon  in  the  lower  base.     These  polygons 


BOOK    VII.  207 

may  be  taken  as  the  bases  of  the  frustum  of  a  right 
pyramid  inscribed  in  the  frustum  of  the  cone. 

]STow,  however  great  the  number  of  sides  of  the  in- 
scribed polygons,  the  convex  surface  of  the  frustum  of 
the  pyramid  is  measured  by  its  slant  height  multiplied  by 
one  half  the  sum  of  the  perimeters  of  its  two  bases, 
(Th.  18) ;  but  when  we  reach  the  limit,  by  making  the 
number  of  sides  of  the  polygon  indefinitely  great,  the 
slant  height,  perimeters  of  the  bases,  and  convex  surface 
of  the  frustum  of  the  pyramid  become,  severally,  the 
slant  height,  circumferences  of  the  bases,  and  convex  sur- 
face of  the  frustum  of  the  cone. 

Hence  the  theorem ;  the  convex  surface  of  the  frustum, 
etc. 

Cor.  1.  If  we  make  oc  —  00,  and,  consequently,  circ. 
oc  =  circ.  0(7,  the  frustum  of  the  cone  becomes  a  cylin- 
der, and  the  half  sum  of  the  circumferences  of  the  bases 
becomes  the  circumference  of  either  base  of  the  cylinder, 
and  the  slant  height  of  the  frustum,  the  altitude  of  the 
cylinder.  Hence,  the  convex  surface  of  a  cylinder  is  meas- 
ured by  the  circumference  of  the  base  multiplied  by  the  alti- 
tude of  the  cylinder. 

Cor.  2.  Kwe  make  oc  =  0,  the  frustum  of  the  cone 
becomes  a  cone.  Hence,  the  convex  surface  of  a  cone  is 
measured  by  the  circumference  of  the  base  multiplied  by  one 
half  the  slant  height  of  the  cone. 

Cor.  3.  If  through  E,  the  middle  point  of  Co,  the  line 
Ff  be  drawn  parallel  to  Oo,  and  Em  perpendicular  to 
Go,  the  line  oc  being  produced,  to  meet  Ff  at/,  we  have, 
because  the  A's  EFC  and  Efc  are  equal, 

w  OC  +  oo 

Em  =  _^-_ . 

If  we  multiply  both  members  of  this  equation  by  2*, 
we  have 

2«.Em  =  2*-0Q+2™. 


208  GEOMETEY. 

that  is,  circ.  Em  is  equal  to  one  half  the  sum  of  the  cir- 
cumferences of  the  two  bases  of  the  frustum.  Hence,  the 
convex  surface  of  the  frustum  of  a  cone  is  measured  by  the 
circumference  of  the  section  made  by  a  plane  half  way  between 
the  two  bases,  and  parallel  to  them,  multiplied  by  the  slant 
height  of  the  frustum. 

Cor.  4.  If  the  trapezoid,  OCco,  be  revolved  about  Oo 
as  an  axis,  the  inclined  side,  Cc,  will  generate  the  con- 
vex surface  of  the  frustum  of  a  cone,  of  which  the  slant 
height  is  Cc,  and  the  circumferences  of  the  bases  are  circ. 
OC  and  circ.  oc.  Hence,  if  a  trapezoid,  one  of  whose  sides 
is  perpendicular  to  the  two  parallel  sides,  be  revolved  about 
the  perpendicular  side  as  an  axis,  it  will  generate  the  frustum 
of  a  cone,  the  inclined  side  opposite  the  axis  generating  the 
convex  surface,  and  the  parallel  sides  the  bases  of  the  frustum. 

THEOREM    XXI. 

The  volume  of  a  cone  is  measured  by  the  area  of  its  base 
multiplied  by  one  third  of  its  altitude. 

Let  V — ABC,  etc.,  be  a  cone;  then 
will  its  volume  be  measured  by  area 
ABC,  etc.,  multiplied  by  \VO. 

Inscribe,  in  the  base  of  the  cone,  any 
regular  polygon,  as  ABCDEF,  which 
may  be  taken  as  the  base  of  a  right  pyra- 
mid, of  which  V  is  the  vertex.  The 
volume  of  this  inscribed  pyramid  will  AJ 
have,  for  its  measure,  (Th.  15), 

polygon  ABCDEF  x  \VO. 

Now,  however  great  the  number  of  sides  of  the  poly- 
gon inscribed  in  the  base  of  the  cone,  it  will  still  hold 
true  that  the  pyramid  of  which  it  is  the  base,  and  whose 
vertex  is  V,  will  be  measured  by  the  area  of  the  poly- 
gon, multiplied  by  one  third  of  VO;  but  when  we 
reach  the  limit,  by  making  the  number  of  sides  indefi- 


BOOK   VII.  209 

nitely  great,  the  polygon  becomes  the  circle  in  which  it 
is  inscribed,  and  the  pyramid  becomes  the  cone. 

Hence  the  theorem ;  the  volume  of  a  cone,  etc. 

Cor.  1.  If  R  denote  the  radius  of  the  base  of  a  cone, 
and  H  its  altitude,  or  axis,  its  volume  will  be  expressed 

by 

hence,  if  Fand  V  designate  the  volumes  of  two  cones, 
of  which  R  and  R '  are  the  radii  of  the  bases,  and  H  and 
Hr  the  altitudes,  we  have 

V:  V  ::  iKx«R2:  ±Hf  x  «Rn  ::  Hx«R2  :  Hf  X  <kR'\ 

From  this  proportion  we  conclude, 

First.  That  cones  having  equal  altitudes  are  to  each  other 
as  their  bases. 

Second.  That  cones  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Cor.  2.  Retaining  the  notation  above,  we  have 

IL       *L       R'2     m 
V  ~   H  x    W 

and,  if  the  two  cones  are  similar, 

Hi  H>  ::  R  :  i2'; 

J27        Rr    ,  E2f        Rf* 

s  =  r;  hence>  M*'^' 

By  substituting  for  the  factors,  in  the  second  member 
of  eq.  ( 1 ),  their  values  successively,  and  resolving  into  a 
proportion,  we  get 

V  :   V  ::  R*  :  R*'; 
and  V  :   V  ::  jEP':  E'\ 

Hence,  similar  cones  are  to  each  other  as  the  cubes  of  the 
radii  of  their  bases,  and  also  as  the  cubes  of  their  altitudes. 

Cor.  3.  A  cone  is  equivalent  to  a  pyramid  having  an  equiv- 
alent base  and  an  equal  altitude. 
18*  o 


210 


GEOMETRY. 


THEOREM    XXII. 

The  volume  of  the  frustum  of  a  cone  is  equivalent  to  the 
sum  of  the  volumes  of  three  cones,  having  for  their  common 
altitude  the  altitude  of  the  frustum,  and  for  their  several 
bases,  the  bases  of  the  frustum  and  a  mean  proportional  be- 
tween them. 

Let  ABQD — abed  be  the  frustum  of  a 
cone ;  then  will  its  volume  be  equiva- 
lent to  the  sum  of  the  volumes,  having 
Oo  for  their  common  altitude,  and  for 
their  bases,  the  circles  of  which,  OG,  oc, 
and  a  mean  proportional  between  00 
and  oc,  are  the  respective  radii. 

Inscribe  in  the  lower  base  of  the  frus- 
tum any  regular  polygon,  and  in  the 
upper  base  a  similar  polygon,  having 
its  sides  parallel  to  those  of  the  first.  These  polygons 
may  be  taken  as  the  bases  of  the  frustum  of  a  right  pyra- 
mid inscribed  in  the  frustum  of  the  cone. 

The  volume  of  the  frustum  of  the  pyramid  is  equiva- 
lent to  the  sum  of  the  volumes  of  three  pyramids,  having 
for  their  common  altitude  the  altitude  of  the  frustum, 
and  for  their  several  bases  the  bases  of  the  frustum,  and 
a  mean  proportional  between  them,  (Th.  16). 

!Now,  however  great  the  number  of  sides  of  the  poly- 
gons inscribed  in  the  bases  of  the  frustum  of  the  cone, 
this  measure  for  the  volume  of  the  frustum  of  the  pyra- 
mid, of  which  they  are  the  bases,  still  holds  true ;  but 
when  we  reach  the  limit,  by  making  the  number  of  the 
sides  of  the  polygon  indefinitely  great,  the  polygons  be- 
come the  circles,  the  frustum  of  the  pyramid  becomes 
the  frustum  of  the  cone,  and  the  three  partial  pyramids, 
whose  sum  is  equivalent  to  the  frustum  of  the  pyramid, 
become  three  partial  cones,  whose  sum  is  equivalent  to 
the  frustum  of  the  cone. 


BOOK    VII.  211 

Hence  the  theorem ;  the  volume  of  the  frustum  of  a  cone,  etc. 

Cor.  1.  Let  R  denote  the  radius  of  the  lower  base,  R  ■' 
that  of  the  upper  base,  and  ^Tthe  altitude  of  the  frustum 
of  a  cone ;  then  will  its  volume  be  measured,  (Th.  21),  by 

±H  x  *R2  +  ±R  x  «Rn  +  i&x  *Rx  Rr, 
since  *R  x  Rr  expresses  the  area  of  a  circle  which  is  a 
mean  proportional  between  the  two  circles,  whose  radii 
are  R  and  Rf. 

E"ow,  if  the  bases  of  the  frustum  become  equal,  or 
R  =  R',  the  frustum  becomes  a  cylinder,  and  each  of  the 
last  two  terms  in  the  above  expression  for  the  volume  of 
the  frustum  of  a  cone  will  be  equal  to  the  first ;  hence, 
the  volume  of  a  cylinder,  of  which  H  is  the  altitude,  and 
R  the  radius  of  the  base,  is  measured  by  H  x  «R2. 

Therefore,  the  volume  of  a  cylinder  is  measured  by  the 
area  of  its  base  multiplied  by  its  altitude. 

Cor.  2.  By  a  process  in  all  respects  similar  to  that  pur- 
sued in  the  case  of  cones,  it  may  be  shown  that  similar 
cylinders  are  to  each  other  as  the  cubes  of  the  radii  of  their 
bases,  and  also  as  the  cubes  of  their  altitudes. 

Cor.  3.  A  cylinder  is  equivalent  to  a  prism  having  an 
equivalent  base  and  an  equal  altitude. 

THEOREM    XXIII. 

If  a  plane  be  passed  through  a  sphere,  the  section  will  be  a 
circle. 

Let  0  be  the  center  of  a  sphere 
through  which  a  plane  is  passed, 
making  the  section  AmBn ;  then 
will  this  section  be  a  circle. 

From  0  let  fall  the  perpendic- 
ular Oo  upon  the  secant  plane, 
and  draw  the  radii  OA,  OB,  and 
Om,  to  the  different  points  in  the 
intersection  of  the  plane  with 
the  surface  of  the  sphere.     Now. 


212  GEOMETRY. 

the  oblique  lines  OA,  OB,  Om,  are  all  equal,  being  radii 
of  the  sphere;  they  therefore  meet  the  plane  at  equal  dis- 
tances from  the  foot  of  the  perpendicular  Oo,  (Cor.,  Th.  4, 
B.VI);  hence  oA,  oB,  om,  etc.,  are  equal:  that  is,  all  the 
points  in  the  intersection  of  the  plane  with  the  surface  of 
the  sphere  are  equally  distant  from  the  point  0.  This 
intersection  is  therefore  the  circumference  of  a  circle  of 
which  o  is  the  center. 

Hence  the  theorem;  if  a  plane  be  passed  through  a 
sphere,  etc. 

Cor.  1.  Since  AB,  the  diameter  of  the  section,  is  a  chord 
of  the  sphere,  it  is  less  than  the  diameter  of  the  sphere ; 
except  when  the  plane  of  the  section  passes  through  the 
center  of  the  sphere,  and  then  its  diameter  becomes  the 
diameter  of  the  sphere.     Hence, 

1.  All  great  circles  of  a  sphere  are  equal. 

2.  Of  two  small  circles  of  a  sphere,  that  is  the  greater 
whose  plane  is  the  less  distant  from  the  center  of  the  sphere. 

3.  All  the  small  circles  of  a  sphere  whose  planes  are  at  the 
same  distance  from  the  center,  are  equal. 

Cor.  2.  Since  the  planes  of  all  great  circles  of  a  sphere 
pass  through  its  center,  the  intersection  of  two  great 
circles  will  be  both  a  diameter  of.  the  sphere  and  a  com- 
mon diameter  of  the  two  circles.  Hence,  two  great  circles 
of  a  sphere  bisect  each  other. 

Cor.  3.  A  great  circle  divides  the  volume  of  a  sphere,  and 
also  its  surface,  equally. 

For,  the  two  parts  into  which  a  sphere  is  divided  by 
any  of  its  great  circles,  on  being  applied  the  one  to  the 
other,  will  exactly  coincide ;  otherwise  all  the  points  in 
their  convex  surfaces  would  not  be  equally  distant  from 
the  center. 

Cor.  4.  The  radius  of  the  sphere  which  is  perpendicular 
to  the  plane  of  a  small  circle,  passes  through  the  center  of  the 
circle. 


BOOK   VII.  213 

Cor.  5.  A  plane  passing  through  the  extremity  of  a  radius 
of  a  sphere,  and  perpendicular  to  it,  is  tangent  to  the  sphere. 

For,  if  the  plane  intersect  the  sphere,  the  section  is  a 
circle,  and  all  the  lines  drawn  from  the  center  of  the 
sphere  to  points  in  the  circumference  are  radii  of  the 
sphere,  and  are  therefore  equal  to  the  radius  which  is  per- 
pendicular to  the  plane,  which  is  impossible,  (Cor.  1,  Th. 
3,  B.  VI).  Hence  the  plane  does  not  intersect  the  sphere, 
and  has  no  point  in  its  surface  except  the  extremity  of 
the  perpendicular  radius.  The  plane  is  therefore  tangent 
to  the  sphere  by  Def  22. 

THEOREM    XXIY. 

If  the  line  drawn  through  the  center  and  vertices  of  two 
opposite  angles  of  a  regular  polygon  of  an  even  number  of 
sides,  be  taken  as  an  axis  of  revolution,  the  perimeter  of  either 
semi-polygon  thus  formed  will  generate  a  surface  whose  measure 
is  the  axis  multiplied  by  the  circumference  of  the  inscribed  circle. 

Let  ABCDEF  be  a  semi-polygon  cut 
off  from  a  regular  polygon  of  an  even 
number  of  sides  by  drawing  the  line  AF 
through  the  center  0,  and  the  vertices  A 
and  F,  of  two  opposite  angles  of  the  poly- 
gon;  then  will  the  surface  generated  by 
the  perimeter  of  this  semi-polygon  re- 
volving about  AF  as  an  axis,  be  meas- 
ured by  AF  X  circumference  of  the  in- 
scribed circle. 

From  m,  the  middle  point,  and  the  extremities  B  and 
0  of  the  side  2? (7,  draw  mn,  BK,  and  OL,  perpendicular  to 
AF;  join  also  m  and  0,  and  draw  BH  perpendicular  to 
CL.  The  surface  of  the  frustum  of  the  cone  generated 
by  the  trapezoid  BKLO,  has  for  its  measure  circ.  mn  X 
BO,  (Cor.  3,  Th.  20).  Since  mO  is  perpendicular  to  BO, 
and  mn  to  BH,  the  two  A's,  BOH  and  mnO,  are  similar, 
and  their  homologous  sides  give  the  proportion 


214  GEOMETRY. 

mn  :  mO  ::  BH  (=  jK£)  :  BO 

and  as  circumferences  are  to  each  other  as  their  radii,  we 
have 

circ.  mn  :  circ.  mO  ::  KL  :  BO 

Hence,     circ.  mn  X  BO  =  circ.  mO  X  KL. 

But  mO  is  the  radius  of  the  circle  inscribed  in  tne 
polygon.  Hence,  the  surface  generated  by  BO  during  the 
revolution  of  the  semi-polygon,  is  measured  by  the  cir- 
cumference of  the  inscribed  circle  multiplied  by  KL,  the 
part  of  the  axis  included  between  the  two  perpendicu- 
lars let  fall  upon  it  from  the  extremities  B  and  0.  The 
surface  generated  by  any  other  side  of  the  semi-polygon 
will  be  measured,  in  like  manner, -by  the  circumference  of 
the  inscribed  circle  multiplied  by  the  corresponding  part 
of  the  axis. 

By  adding  the  measures  of  the  surfaces  generated  by 
the  several  sides  of  the  semi-polygon,  we  get 

Circ.  mO  x  {AK  +  KL  +  LN +  NM+  MF) 

for  the  measure  of  the  whole  surface. 

Hence  the  theorem ;  if  the  line  drawn  through  the  cen- 
ter, etc. 

Oor.  It  is  evident  that  the  surface  generated  by  any 
portion,  as  OB  and  BF,  of  the  perimeter,  is  measured  by 
circ.  mO  x  LM. 

THEOREM    XXV. 

The  surface  of  a  sphere  is  measured  by  the  circumference 
of  one  of  its  great  circles  multiplied  by  its  diameter. 

Let  a  sphere  be  generated  by  the  revolution  of  the 
semi-circle,  ARF,  about  its  diameter,  AF;  then  will  the 
surface  of  the  sphere  be  measured  by 

Circ.  AO  x  AF. 

Inscribe  in  the  semi-circle  any  regular  semi-polygon, 
and  let  it  be  revolved,  with  the  semi-circle,  about  the  axis 


BOOK  VII.  215 

AF;  the  surface  generated  by  its  perim- 
eter will  be  measured  by 

Circ.  mOx  AF,  (Th.  24), 

and  this  measure  will  hold  true,  how- 
ever great  the  number  of  sides  of  the  in-  H| 
scribed  semi-polygon.  But  as  the  num- 
ber of  these  sides  is  increased,  the 
radius  mO,  of  the  inscribed  semi-circle, 
increases  and  approaches  equality  with 
the  radius,  AO;  and  when  we  reach  the  limit,  by 
making  the  number  of  sides  indefinitely  great,  the  radii 
and  semi-circles  become  equal,  and  the  surface  generated 
by  the  perimeter  of  the  inscribed  semi-polygon  becomes 
the  surface  of  the  sphere.  Therefore,  the  surface  of  the 
sphere  has,  for  its  measure, 

Girc.  AO  x  AF. 

Hence  the  theorem ;  the  surface  of  a  sphere  is  meas- 
ured, etc. 

Cor.  1.  A  zone  of  a  sphere  is  measured  by  the  circumfer- 
ence of  a  great  circle  of  the  sphere  multiplied  by  the  altitude 
of  the  zone. 

For,  the  surface  generated  by  any  portion,  as  CD  and 
DF,  of  the  perimeter  of  the  inscribed  semi-polygon  has, 
for  its  measure,  circ.  mO  X  LM,  (Cor.  Th.  24) ;  and  as 
the  number  of  the  sides  of  the  semi-polygon  increases, 
LM  remains  the  same,  the  radius  mO  alone  changing, 
and  becoming,  when  we  reach  the  limit,  equal  to  AO) 
hence,  the  surface  of  the  zone  is  expressed  by 

Circ.  AO  x  LM, 
whether  the  zone  have  two  bases,  or  but  one. 

Cor.  2.  Let  H  and  H'  denote  the  altitudes  of  two 
zones  of  spheres,  whose  radii  are  R  and  R ' ;  then  these 
zones  will  be  expressed  by  2*R  x  H  and  2*R '  x  Hf; 
and  if  the  surfaces  of  the  zones  be  denoted  by  Z  and  Zr, 
we  have 


216  GEOMETRY. 

Z  :  Z'  : :  2«R  x  E  :  2«Rf  x  E'  : :  R  x  E  :  Rf  x  E'. 

Hence,  1.  Zones  in  different  spheres  are  to  each  other  as 
their  altitudes  multiplied  by  the  radii  of  the  spheres. 

2.  Zones  of  equal  altitudes  are  to  each  other  as  the  radii 
of  the  spheres. 

3.  Zones  in  the  same,  or  equal  spheres,  are  to  each  other  as 
their  altitudes. 

Cor.  3.  Let  R  denote  the  radius  of  a  sphere;  then  will 
its  diameter  be  expressed  by  2R,  and  the  circumference 
of  a  great  circle  by  2*R ;  hence  its  surface  will  be  ex- 
pressed by 

2«R  x2E  =  ±«R\ 

That  is,  the  surface  of  a  sphere  is  equivalent  to  the  area  of 
four  of  its  great  circles. 

Cor.  4.  The  surfaces  of  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

THEOREM    XXVI. 

If  a  triangle  be  revolved  about  either  of  its  sides  as  an  axis, 
the  volume  generated  will  be  measured  by  one  third  of  the  prod- 
uct of  the  axis  and  the  area  of  a  circle,  having  for  its  radius 
the  perpendicular  let  fall  from  the  vertex  of  the  opposite 
angle  on  the  axis,  or  on  the  axis  produced. 

First.  Let  the  triangle  ABC, 
in  which  the  perpendicular  from 
C  falls  on  the  opposite  side,  AB, 
be  revolved  about  AB  as  an  axis; 
then  will  *Yol.  A  ABC  have,  for 
its  measure,  %AB  x  *CD . 

The  two  A's  into  which  A  ABC  is  divided  by  the 
perpendicular  DC,  are  right-angled,  and  during  the  rev- 
olution 'they  will  generate  two  cones,  having  for  their 

*  Vol.  A  ABC,  cone  A  ABC,  are  abbreviations  for  volume  gener- 
ated by  A  ABC,  cone  generated  by  A  ABC',  and  surfaces  of  revolu- 
tion generated  by  lines  will  hereafter  be  denoted  by  like  abbreviations. 


BOOK    VII.  217 

common  base  the  circle,  of  which  DO  is  the  radius,  and 
for  their  axes  the  parts  DA  and  DB,  into  which  AB  is 
divided. 


Now,  *Cone  A  ADO  is  measured  by  \AD  x  mj>(jy 
(Th.  21),  and  cone  A  BDO,  by  ±BD  x  iDQ* ;  but  these 
two  cones  compose  Yol.  A  ABO;  and  by  adding  their 
measures,  we  have,  for  that  of  Yol.  A  ABO, 

iAD  x  *D02  +  iBD  x  r~D02  =  \AB  x  <HJC\ 
Second.  Let  the  trian- 
gle EFGr,  in  which  the 
perpendicular  from  Gr 
falls  on  the  opposite  side 
EF produced,  be  revolved 
about  EF  as  an  axis ; 
then  will  Yol.   A  FFa     E  F  G 

have,  for  its  measure,  ^EF  x  <kGtH\  CrH  being  the  per- 
pendicular on  EF  produced.  For,  in  this  case  it  is  appa- 
rent, that  Yol.  A  EFGr  is  the  difference  between  the 
cone  A  ERG-  and  the  cone  A  FHGr.  The  first  cone  has, 
for  its  measure,  \EH  x  «GrE\  and  the  second,  for  its 
measure,  ^FH  x  irGfH2 ;  hence,  by  subtraction,  we  have 

Vol.  A  EFG  =  %EH  X  *  GH2  —  \FH  X  nGH2  =  %EF  X  7i~GH2. 
Hence  the  theorem ;  if  a  triangle  be  revolved  about  either 
of  its  sides,  etc. 

Scholium. — If  we  take  either  of  the  above  expressions  for  the  meas- 
ure of  the  volume  generated  by  the  revolution  of  a  triangle  about  one 
of  its  sides,  for  example  the  last,  and  factor  it  otherwise,  we  have 

iEFX  7tGH2  =  FFx^GHxi7tx2GS=FFxiGHx  ^ 

j 

Now,  EF  X  %GH  expresses  the  area  of  the  triangle  EFG;  and 

2rt  X  GH  .  . 

,  one  third  of  the  circumference  described  by  the.  point  Q 

o 

during  the  revolution. 

The  expression,  IAB  X  TtDC2,  maybe  factored  and  interpreted  in  the 

*  See  note  on  the  preceding  page. 

19 


218 


GEOMETEY. 


same  manner.  Hence,  we  conclude  that  the  volume  generated  by  the 
revolution  of  a  triangle  about  either  of  its  sides,  is  measured  by  the  area 
of  the  triangle  multiplied  by  one  third  of  the  circumference  described  in 
the  revolution  by  the  vertex  of  the  angle  opposite  the  axis. 


THEOREM    XXVII. 

The  volume  generated  by  the  revolution  of  a  triangle  about 
any  line  lying  in  its  plane,  and  passing  through  the  vertex  of 
one  of  its  angles,  is  measured  by  the  area  of  the  triangle  mul- 
tiplied by  two  thirds  of  the  circumference  described,  in  the 
revolution,  by  the  middle  'point  of  the  side  opposite  the  vertex 
through  which  the  axis  passes. 

Let  the  triangle  ABO  be 
revolved  about  the  line 
AG,  drawn  through  the 
vertex  A,  and  lying  in  the 
plane  of  the  triangle,  and 
let  HE  be  the  perpendicu- 
lar let  fall  from  H,  the 
middle  point  of  BO,  upon 
the  axis  AG ;  then  will  Vol.  a  ABO  have,  for  its  meas- 
ure, A  ABO  X  §  circ.  HE. 

From  the  extremities  of  BO,  let  fall  the  perpendicu- 
lars BE  and  OB,  on  the  axis;  and  from  A  draw  AK per- 
pendicular to  BO,  or  BO  produced,  and  produce  OB, 
until  it  meets  the  axis  in  G. 

]STow,  it  is  evident  that  Yol.  A  ABO  is  the  difference 
between  Yol.  A  AGO  and  Yol.  A  AGB.  But  Yol. 
A  AGO  is  expressed  by  a  A  GO  X  J  circ.  OB;  and  Yol. 
A  AGB,  by  A  AGB  x  J  circ.  BE,  (Scholium,  Th.  26). 
Hence, 

Vol.  A  ABC  ==  A  AGC  X  \  circ.  CD  —  A  AGB  X  £  circ.  BF. 

Substituting  for  areas  of  A's,  and  for  circumferences, 
their  measures,  we  have 


BOOK   VII.  219 

Vol.  A  ABC=  GC  X  \AK  X  ^hfR—  GB  X  \AK  X  ^M 

o  o 

=  GCx  UKX  2^£R--{GC--BC)  X  IAKX  &%H 

^GCXUKX2-^^-—  GCxUKX—~  +  BCxUKX^^- 
o  o  o 

=  GCX  \AK  X  ~{CJ)  —  5Z)  +  5C  X  M^  X  =~i 

3  o 

But  jSiV' being  drawn  parallel  to  AG,  we  have 
CJSF  =   CD  —  BF; 
hence,  substituting  this  value  for  CD  —  BF,  in  the  first 
term  of  the  second  member  of  the  last  equation,  we  have 

Yol.AABO=aOxiAKx^4^+BOxiAKx  2 


=  GCx  OJSTx  lAKx  **  +  BCx  \AKx- 


3 

~  +  BCx  \AKx  — g- 

by  changing  the  order  of  factors  in  the  first  term  of  the 
second  member.  The  homologous  sides  of  the  similar 
triangles,  GOD  and  BON,  give  the  proportion 

GO  :  OB  : :  BO  :  ON 
whence,  GCx  ON  =  CD  x  BC 

Substituting  this  value  for  GO  x  ON,  in  the  last  equa- 
tion above,  and  arranging  the  factors  as  before,  it  becomes 

Zx.CD  ,  ™.    ,  ,_.    2«.BF 
3 


Vol.  A  ABC=  BC  x  i AKx  =I£±:  +  BC x  \AK x 


=  BCxiAKx2^0A±-B-^. 
ButCD  +  BF  =  2HF;  hence 

Vol.  A  ABC=BCx  lAKx  —~=BCx  \AKx  §.2«.HF; 

o 

and  since 

BC  x  \AK=  A  ABC,  and  $  x  2«.HF  =  }  circ.  BF, 
this  measure  conforms  to  the  enunciation. 

It  only  remains  for  us  to  consider  the  case  in  which 
the  axis  is  parallel  to  the  base  BC  of  the  triangle.     The 


220 


GEOMETRY. 


preceding  demonstration  will  not  now  apply,  because  it 
supposes  BO,  or  BO  produced,  to  intersect  the  axis. 

Let  the  axis  AE,  be  parallel  to  the 
base  BO,  of  the  A  ABO.  From  B 
and  0  let  fall  on  the  axis  the  perpen- 
diculars BE  and  CD. 

Now  it  is  plain  that 

Vol.  A  ABO  =  cylinder  rectangle  BODE  + 
cone  a  ADO —  cone  A  AEB. 

Substituting  in  second  member,  for  cylinder  and  cones, 
their  measures,  we  have 

Vol.  AABC=  DE  x  *OD2  +  %AD  x  «~OD2—  \AE  x  «BE2 
=§DEx  *CD2+\DEx  «OD2+§ADx«CD2--iAEx  *BE\ 

But  BE  =  CD,  and  \DE  +  \AD  =  \AD.  Reducing  by 
these  relations,  we  have 

Yol.  a  ABC=  %DE  x  «CD2=  \DE  x  \OD  x  4*.OZ> 
_  DE  x  \ODx  %.<L«.OD  =  BO  x  \OD  x  %.2«.OD. 

And,  since  BO  x  %CD  expresses  the  area  of  the  tri- 
angle ABO,  and  %.2<x.CD,  two  thirds  of  the  circumfer- 
ence described  by  any  point  of  the  base,  this  expression 
also  conforms  to  the  enunciation. 

Hence  the  theorem ;  the  volume  generated  by  the  revolu- 
tion, etc. 

Oor.  If  the  generating 
triangle  becomes  isosceles, 
the  perpendicular  from  A 
meets  the  base  at  its  middle 
point.  In  this  case,  if  we 
resume  the  expression 

BOx  lAKx  — x 

o 

it  becomes 

BO  x  \AK  x  KE  x  J« 


/  9 

J 

K 

B 

A              I 

)         J 

1         I 

? 

BOOK   VII.  221 

But,  since  AKis  perpendicular  to  BO,  and  KB  to  BJSF, 
the  a's  A  KB  and  CBN  are  similar,  and  their  homolo- 
gous sides  give  the  proportion 

BO  :  BJST  ::  AK  :  KB 

whence,  BCxKB  =  .Bi^x  JLJ5T 

Changing  the  order  of  factors  in  the  last  expression  on 
the  preceding  page,  and  replacing  BOxKE  by  its  value, 
it  becomes 

\AKx  AKxBJSTx  ]i  =  AK2  x  BN  x  |* 

Hence, 

Vol.  A^LB(7=f*'  x  33?  x  J9iV.  =  f*  xAK2  xJDF 

That  is,  £A<?  volume  generated  by  the  revolution  of  an  isos- 
celes about  any  line  drawn  through  its  vertex  and  lying  in  the 
plane  of  the  triangle,  is  measured  by  %«  times  the  square  of 
the  perpendicular  of  the  triangle  multiplied  by  the  part  of  the 
axis  included  between  the  two  perpendiculars  let  fall  upon  it 
from  the  extremities  of  the  base  of  the  triangle. 

'  Scholium.  —  If  we  resume  the  equation 

Vol.  A  ABC  =  BC  X  \AK  X  ^™ 

o 

and  change  the  order  of  the  factors  in  the  second  member,  it  may  be 
put  under  the  form 

Vol.  A  ABC  =  BC  X  2h.HE  X  \AK. 

But  during  the  revolution  of  the  triangle,  the  side  BC  generates  the 
surface  of  the  frustum  of  a  cone,  which  surface  has  for  its  measure 

BC  X  2rt.HE  (Th.  20,  Cor.  3). 

Hence,  the  above  equation  may  be  thus  interpreted :  The  volume 
generated  by  the  revolution  of  a  triangle  about  any  line  lying  in  its  plane 
and  passing  through  the  vertex  of  one  of  its  angles,  is  measured  by  the 
surface  generated,  during  the  revolution,  by  the  side  opposite  the  vertex 
through  which  the  axis  passes  multiplied  by  one  third  of  the  perpen- 
dicular drawn  from  the  vertex  to  that  side. 


19* 


222 


GEOMETRY. 


THEOREM    XXVIII. 


If  the  line  drawn  through  the  center  and  vertices  of  two  op- 
posite  angles  of  a  regular  polygon,  of  an  even  number  of 
sides,  be  taken  as  an  axis  of  revolution,  either  semi-polygon 
thus  formed  will,  during  this  revolution,  generate  a  volume 
which  has,  for  its  measure,  the  surface  generated  by  the 
perimeter  of  the  semi-polygon  multiplied  by  one  third  of  its 
apothem. 

Let  ABODE  be  a  regular  semi-poly- 
gon, cut  off  from  a  regular  polygon 
of  an  even  number  of  sides,  by  draw- 
ing a  line  through  the  center,  0,  and 
the  vertices,  A  and  E,  of  two  opposite 
angles  of  the  polygon ;  then  will  the 
volume  generated  by  the  revolution 
of  this  semi-polygon  about  AE,  as  an 
axis,  be  measured  by  (Sur.  AB  -f  sur. 
BO  +  sur.  CD  -f  sur.  DE)  x  JOm,  Om 
being  the  apothem  of  the  polygon. 

For,  if  from  the  center  of  0,  the  lines  OB,  00,  OB,  be 
drawn  to  the  vertices  of  the  several  angles  of  the  semi- 
polygon,  it  will  be  divided  into  equal  isosceles  triangles, 
the  perpendicular  of  each  being  the  apothem  of  the 
polygon. 

Now,  the  volume  generated  by  A  AOB  has,  for  its 
measure, 

Sur.  AB  x  \0m, 

that  by  A  BOO,  Sur.  BO  x  \0m, 
"  A  OOB,  Sur.  OB  x  \0m, 
"      A  DOE,    Sur.  DE  x  %0m,  (Scholium,  Th.  27). 

By  the  addition  of  the  measures  of  these  partial  vol- 
umes, we  find,  for  that  of  the  whole  volume, 

Vol.  semi-polygon  ABODE  =  sur.  perimeter  ABODE  X  \Om, 

and  were  the  number  of  the  sides  of  the  semi-polygon 


BOOK  VII.'  223 

increased  or  diminished,  the  reasoning  would  be  in  no 
wise  changed. 

Hence  the  theorem ;  if  the  line  drawn  through  the  cen- 
ter, etc. 

Scholium. — The  volume  generated  by  any  portion  of  the  semi-poly- 
gon, as  that  composed  of  the  two  isosceles  a's  DOC,  COD,  is  meas- 
ured by 

Sur.  perimeter  BCD  X  10m. 

THEOREM    XXIX. 

The  volume  of  a  sphere  is  measured  by  its  surface  multi- 
plied by  one  third  of  its  radius. 

Let  a  sphere  be  generated  by  the 
revolution  of  the  semicircle  AOE, 
about  its  diameter,  AE,  as  an  axis; 
then  will  the  volume  of  the  sphere  be 
measured  by 

sur.  semi-circ.  OA  x  %OA. 

For,  inscribe  in  the  semi-circle  any 
regular  semi  -  polygon,  as  ABODE, 
and  let  it,  together  with  the  semi-cir- 
cle, revolve  about  the  axis  AE.  The 
semi-polygon  will  generate  a  volume  which  has,  for  its 
measure, 

Sur.  perimeter  ABODE  x  \Om,  (Th.  28), 

in  which  Om  is  the  apothem  of  the  polygon. 

Now,  however  great  the  number  of  sides  of  the  in- 
scribed regular  semi-polygon,  this  measure  for  the  volume 
generated  by  it,  will  hold  true ;  but  when  we  reach  the 
limit,  by  making  the  number  of  sides  indefinitely  great, 
the  perimeter  and  apothem  become,  respectively,  the 
semi-circumference  and  its  radius,  and  the  volume  gen- 
erated by  the  semi-polygon  becomes  that  generated  by 
the  semi-circle,  that  is,  the  sphere.     Therefore, 

Vol.  sphere  =  sur.  semi-circ.  OA  x  \OA. 


224 


GEOMETRY. 


Scholium  1. — If  we  take  any  portion  of  the  inscribed  semi-polygon, 
as  BOC,  the  volume  generated  by  it  is  measured  by  sur.  BC  X  $Om, 
(Scholium,  Th.  27) ;  and  when  we  pass  to  the  limit,  this  volume  be- 
comes a  sector,  and  sur.  BC  &  zone  of  the  sphere,  which  zone  is  the 
base  of  the  sector.  Hence,  the  volume  of  a  spherical  sector  is  measured 
by  the  zone  which  forms  its  base  multiplied  by  one  third  of  the  radius 
of  the  sphere. 

Scholium  2. —  Let  R  denote  the  radius  of  a  sphere;  then  will  its 
diameter  be  represented  by  2R.  Now,  since  the  surface  of  a  sphere  is 
equivalent  to  the  area  of  four  of  its  great  circles,  and  the  area  of  a 
great  circle  is  expressed  by  7tR2,  we  have 

Vol.  sphere  =  4hR  2  X^  =  {jtR3. 

And  since  R3  —  |( 2R )3,  we  also  have 

Vol.  sphere  =  &R3  =  %rt{2R)s. 

That  is,  the  volume  of  a  sphere  is  measured  four  thirds  of*  times  the 
cube  of  the  radius,  or  by  one  sixth  of  7t  times  the  cube  of  the  diameter. 

THEOREM    XXX. 

The  surface  of  a  sphere  is  equivalent  to  two  thirds  of  the 
surface,  bases  included,  and  the  volume  of  a  sphere  to  two 
thirds  of  the  volume,  of  the  circumscribing  cylinder. 

Let  AMD  be  a  semi-circle,  and 
ABQD  a  rectangle  formed  by 
drawing  tangents  through  the 
middle  point  and  extremities  of 
the  semi-circumference,  and  let  M 
the  semi-circle  and  rectangle  be 
revolved  together  about  AD  as 
an  axis.  The  rectangle  will  thus  ( 
generate  a  cylinder  circumscribed 
about  the  sphere  generated  by  the  semi-circle. 

First.  The  diameter  of  the  base,  and  the  altitude  of 
the  cylinder,  are  each  equal  to  the  diameter  of  the 
sphere ;  hence  the  convex  surface  of  the  cylinder,  being 
measured  by  the  circumference  of  its  base  multiplied  by 
its  altitude,  (Cor.  1,  Th.  20),  has  the  same  measure  as 
the  surface  of  the  sphere,  (Th.  25).  But  the  surface  of 
the  sphere  is  equivalent  to  four  great  circles,  (Cor.  3, 


BOOK   VII.  225 

Th.  25).  Hence,  the  convex  surface  of  the  cylinder  is 
equivalent  to  four  great  circles ;  and  adding  to  these  the 
bases  of  the  cylinder,  also  great  circles,  we  have  the 
whole  surface  of  the  cylinder  equivalent  to  six  great 
circles.  Therefore,  the  surface  of  the  sphere  is  four 
sixths  =  two  thirds  of  the  surface  of  the  cylinder,  in- 
cluding its  bases. 

Second.  The  volume  of  the  cylinder,  being  measured 
by  the  area  of  the  base  multiplied  by  the  altitude,  (Cor. 
1,  Th.  22),  is,  in  this  case,  measured  by  the  area  of  a 
great  circle  multiplied  by  its  diameter  =  four  great  cir- 
cles multiplied  by  one  half  the  radius  of  the  sphere. 

But  the  volume  of  the  sphere  is  measured  by  four 
great  circles  multiplied  by  one  third  of  the  radius,  (Scho- 
lium 2,  Th.  29).     Therefore, 

Vol.  sphere  :  Vol.  cylinder  : :  J  :  J  : :  2  :  3 ; 

whence,     Vol.  sphere  =  §  Vol.  cylinder. 

Hence  the  theorem ;  the  surface  of  a  sphere  is  equiva- 
lent, etc, 

,  Cor.  The  volume  of  a  sphere  is  to  the  volume  of  the  cir- 
cumscribed cylinder,  as  the  surface  of  the  sphere  is  to  the  sur- 
face of  the  cylinder. 

Scholium. — Any  polyedron  circumscribing  a  sphere,  may  be  regarded 
as  composed  of  as  many  cones  as  the  polyedron  has  faces,  the  center  of 
the  sphere  being  the  common  vertex  of  these  cones,  and  the  several 
faces  of  the  polyedron  their  bases.  The  altitude  of  each  cone  will  be 
a  radius  of  the  sphere ;  hence  the  volume  of  any  one  cone  will  be 
measured  by  the  area  of  the  face  of  the  polyedron  which  forms  its 
base,  multiplied  by  one  third  of  the  radius  of  the  sphere.  There- 
fore, the  aggregate  of  these  cones,  or  the  whole  polyedron,  will  be 
measured  by  the  surface  of  the  sphere  multiplied  by  one  third  of  the 
radius  of  the  sphere. 

But  the  volume  of  the  sphere  is  also  measured  by  the  surface  of  the 
sphere  multiplied  by  one  third  of  its  radius.     Hence, 

Sur.  polyedron  :  Sur.  sphere  : :  Vol.  polyedron  :  Vol.  sphere. 

That  is,  the  surface  of  any  circumscribed  polyedron  is  to  the  surface 
of  the  sphere,  as  the  volume  of  the  polyedron  is  to  the  volume  of  the 
sphere. 


226  GEOMETRY. 

THEOREM  XXXI. 

The  volume  generated  by  the  revolution  of  the  segment  of  a 
circle  about  a  diameter  of  the  circle  exterior  to  the  segment,  is 
measured  by  one  sixth  of  «r  times  the  square  of  the  chord  of 
the  segment,  multiplied  by  the  part  of  the  axis  included  be- 
tween the  perpendiculars  let  fall  upon  it  from  the  extremities 
of  the  chord. 

Let  BOB  be  a  segment  of  the  circle, 
whose  center  is  0,  and  AH  a  part  of  a 
diameter  exterior  to  the  segment.  Draw 
the  chord  BB,  and  from  its  extremities 
let  fall  the  perpendiculars,  BF,  BF  on 
AH;  also  draw  Om  perpendicular  to 
BD.  The  spherical  sector  generated 
by  the  revolution  of  the  circular  sector 
BCJDO  about  AH,  is  measured  by  zone  BB  x  iBO, 
(Scholium  1,  Th.  29),  =  2«.BO  x  FF  x  %B0  =  §  <^BOl  x 
FF;  and  the  volume  generated  by  the  isosceles  triangle 
BOB  is  measured  by 

&0m   x  FF,  (Cor.  1,  Th.  27). 
The  difference  between  these  two  volumes  is  that  gen- 
erated by  the  circular  segment  BOB,  which  has,  there- 
fore, for  its  measure, 


%«FF{BO  —  Om)  =  l«FF  x  Bm,  (Th.  39,  B.  I). 

But  since  Bm  =  IBB,  'Bm2  =  \BB* ;  hence,  by  sub- 
stituting, we  have 

Yol.  segment  BOB  =  f  *FF  x  %BB2  m  \v~BD2  x  FF. 
Hence  the  theorem. 

THEOREM    XXXII. 

The  volume  of  a  segment  of  a  sphere  has,  for  its  measure, 
the  half  sum  of  the  bases  of  the  segment  multiplied  by  its  alti- 
tude, plus  the  volume  of  a  sphere  which  has  this  altitude  for 
its  diameter. 


BOOK   VII.  227 

Let  BOB  be  the  arc  of  a  circle,  and 
BF  and  BE  perpendiculars  let  fall 
from  its  extremities  upon  a  diameter,  q^ 
of  which  AH  is  a  part ;  then,  if  the 
area  BCBEF  be  revolved  about  AH  r 
as  an  axis,  a  spherical  segment  will 
be  generated,  for  the  volume  of  which 
it  is  proposed  to  find  a  measure. 

The  circular  segment  will  generate  a  volume  meas- 
ured by  \*BB*  x  EF,  (Th.  31) ;  and  the  frustum  of  the 
cone  generated  by  the  trapezoid  BBEF  will  have,  for 
its  measure, 

\^BF2  xEF+  \«~BE2  xEF+  ^BF  xBEx  EF,  (Th.  22), 
=  i«EF(BF2  +  ~BE2  +  BF  x  BE). 

But  the  sum  of  these  two  volumes  is  the  volume  of 
the  spherical  segment,  which  has,  therefore,  for  its 
measure, 

i*EF  (BB*  +  2BF2  +  2BE2  +  2BF  x  BE) 

,  From  B  let  fall  the  perpendicular  Bn  on  BE;  then  will 

Bn  =  BE—nE  =  BE—  BF; 

hence,      Bn  =  BE*  —  2BE  x  BF  +  BF2 ; 

and  since    BB2  =  Bn2  +  ~Bn    =  EF2  +  ~Bn, 

we  have    BB2  =  EF2 +  BE2 +  BF2 —  2BE  x  BF. 

By  substituting  this  value  for  J5D2,  in  the  above  meas- 
ure for  the  volume  of  the  segment,  we  find 


btEF(EF'  +  DE'+BF  —2DExBF+2BF'+  2DE  +2BFXDE) 


2    .     o-F^»2\  1    .^^3    ,      ^^(itDE    +7tBF 


\*EF  {EF'+ZDE'+ZBF')  =  faEF*  +  EF 


■). 


2 

"Which  last  expression  conforms  to  the  enunciation. 
Hence  the  theorem ;  the  volume  of  a  segment  of  a  sphere, 

etc. 

Cor.  When  the  segment  has  but  one  base,  BF  becomes 

zero,  and  EF  becomes  EA;   and  the  final  expression 


GEOMETRY. 

which  we  found  for  the  volume  of  the  segment  reduces 

to  

«DE2 


%*JEA3  +  EA   x 


2 

Hence,  A  spherical  segment  having  but  one  base,  is  equiva- 
lent to  a  sphere  whose  diameter  is  the  altitude  of  the  segment, 
'plus  one  half  of  a  cylinder  having  for  base  and  altitude  the 
base  and  altitude  of  the  segment. 

Scholium. — "When  the  spherical  segment  has  a  single  base,  we  may 

put  the  expression,  \rtEA   +  EA  X  — — ,  under  a  form  to  indicate  a 

convenient  practical  rule  for  computing  the  volume  of  the  segment. 

Thus,  since  the  triangle  DEO  is  right-angled,  and  OE=  OA  —  EA, 
we  have 


DE'  =DO  —OE=  OA  —  OA'  +  20A  X  EA  —  EA 
=  20A  X  EA—  ~EA2. 
By  substituting  this  value  for  BE2  in  the  expression  for  the  volume 
of  the  segment,  we  find 

UEA6  +  EA  X  %  X  {20A  X  EA—~EJl) 

A 

=  \^EA*  -f  EA2  X  %  {20A  —  EA) 

^=\*EA*  ■\-\1tZEA\20A  —  EA) 
=  I  xEA  \EA  +  6.0^  —  SEA) 
=  %7tEA2{6.0A  —  2EA) 
=  $7tEA2{Z0A  —  EA) 

Hence,  the  volume  of  a  spherical  segment,  having  a  single  base,  is 
measured  by  one  third  of  rt  times  the  square  of  the  altitude  of  the  seg- 
ment, multiplied  by  the  difference  between  three  times  the  radius  of  the 
sphere  and  this  altitude. 

RECAPITULATION 

Of  some  of  the  principles  demonstrated  in  this  and  the  pre- 
ceding Books. 

Let  J?  denote  the  radius,  and  D  the  diameter  of  any 
circle  or  sphere,  and  H  the  altitude  of  a  cone,  or  of  a 
segment  of  a  sphere ;  then, 


BOOK  VII.  229 


|  =  2*R  x  S. 


or, 


Circumference  of  a  circle        =  2nR. 
Surface  of  a  sphere  =  4*i22. 

Zone  forming  the  base  of  a 

segment  of  a  sphere, 
Volume  or  solidity  of  a  sphere  =  i*R3,  or  J*!)3. 
Volume  of  a  spherical  sector  =  f  *jR2  x  H. 
Volume  of  a  cone,  of  which  ^ 

R  is   the  radius  of  the  I  =  ^R2  x  E. 

base  J 

Volume  of  a  spherical  seg-^ 

ment,  of  which  R'  is  the 

radius  of  one  base,  and 

R"   the   radius    of   the 

other,  and  whose  altitude 

is  JET, 
If  the  segment  has  but  one  ^     _  t    ,~3       jj-rtR'2 

base,  i2"  =  zero,  and  the  I  ~~  ¥*       +     ,~2~  ' 

volume  of  the  segment,  J   =  J*iP(372  —  jff). 

PRACTICAL    PROBLEMS. 

1.  The  diameter  of  a  sphere  is  12  inches  ;  how  many 
cubic  inches  does  it  contain  ?  Ans.  904.78  cu.  in. 

2.  What  is  the  solidity  of  the  segment  of  a  single  base 
that  is  cut  from  a  sphere  12  inches  in  diameter,  the  altitude 
of  the  segment  being  3  inches?       Ans.  141.371  cu.  in. 

3.  The  surface  of  a  square  is  68  square  feet ;  what  is 
its  diameter  ?  Ans.  D  =  4.625  feet. 

4.  If  from  a  sphere,  whose  surface  is  68  square  feet,  a 
segment  be  cut,  having  a  depth  of  two  feet  and  a  single 
base,  what  is  the  convex  surface  of  the  segment  ? 

Ans.  29.229+  sq.  ft. 

5.  What  is  the  solidity  of  the  sphere  mentioned  in  the 
two  preceding  examples,  and  what  is  the  solidity  of  the 
segment,  having  a  depth  of  two  feet,  and  but  one  base  ? 

a         (  Solidity  of  sphere,     52.71  cu.  in. 
\        "    %    "   segment,  20.85       " 
20 


280  GEOMETRY. 

6.  In  a  sphere  whose  diameter  is  20  feet,  what  is  the 
solidity  of  a  segment,  the  bases  of  which  are  on  the  same 
side  of  the  center,  the  first  at  the  distance  of  3  feet  from 
it,  and  the  second  of  5  feet ;  and  what  is  the  solidity  of 
a  second  segment  of  the  same  sphere,  whose  bases  are 
also  on  the  same  side  of  the  center,  and  at  distances 
from  it,  the  first  of  5  and  the  second  of  7  feet  ? 

a       (  Solidity  of  first  segment,  525.7  cu.  ft. 
I        "        "  second    "        400.03     " 

7.  If  the  diameter  of  the  single  base  of  a  spherical 
segment  be  16  inches,  and  the  altitude  of  the  segment  4 
inches,  what  is  its  solidity  ?  * 

Ans.  435.6352  cubic  inches. 

8.  The  diameter  of  one  base  of  a  spherical  segment  is 
18  inches,  and  that  of  the  other  base  14  inches,  these 
bases  being  on  opposite  sides  of  the  center  of  the  sphere, 
and  the  distance  between  them  9  inches ;  what  is  the 
volume  of  the  segment,  and  the  radius  of  the  sphere  ? 

a         f  Vol.  seg.,  2600.3  cubic  inches. 
\  Rad.  of  sphere,  9.4027  inches. 

9.  The  radius  of  a  sphere  is  20,  the  distance  from  the 
center  to  the  greater  base  of  a  segment  is  10,  and  the 
distance  from  the  same  point  to  the  lesser  base  is  16 ; 
what  is  the  volume  of  the  segment,  the  bases  being  on 
the  same  side  of  the  center  ?  Ans.  4297.7088. 

10.  If  the  diameter  of  one  base  of  a  spherical  segment 
be  20  miles,  and  the  diameter  of  the  other  base  12  miles, 
and  the  altitude  of  the  segment  2  miles,  what  is  its 
solidity,  and  what  is  the  diameter  of  the  sphere  ? 

*  First  find  the  radius  of  the  sphere. 


BOOK   VIII.  231 


BOOK  VIII 


PRACTICAL  GEOMETRY. 

APPLICATION    OF    ALGEBRA   TO   GEOMETRY,   AND    ALSO 
PROPOSITIONS  FOR  ORIGINAL  INVESTIGATION. 

No  definite  rules  can  be  given  for  the  algebraic  solu- 
tion of  geometrical  problems.  The  student  must,  in  a 
a  great  measure,  depend  on  his  own  natural  tact,  and 
his  power  of  making  a  skillful  application  of  the  geomet- 
rical and  analytical  knowledge  he  has  thus  far  obtained. 

The  known  quantities  of  the  problem  should  be  repre- 
sented by  the  first  letters  of  the  alphabet,  and  the  un- 
known by  the  final  letters ;  and  the  relations  between 
these  quantities  must  be  expressed  by  as  many  inde- 
pendent equations  as  there  are  unknown  quantities.  To 
obtain  the  equations  of  the  problem,  we  draw  a  figure, 
the  parts  of  which  represent  the  known  and  unknown 
magnitudes,  and  very  frequently  it  will  be  found  neces- 
sary to  draw  auxiliary  lines,  by  means  of  which  we  can 
deduce,  from  the  conditions  enunciated,  others  that  can 
be  more  conveniently  expressed  by  equations.  In  many 
cases  the  principal  difficulty  consists  in  finding,  from  the 
relations  directly  given  in  the  statement,  those  which 
are  ultimately  expressed  by  the  equations  of  the  problem. 
Having  found  these  equations,  they  are  treated  by  the 
known  rules  of  algebra,  and  the  values  of  the  required 
magnitudes  determined  in  terms  of  those  given. 


232 


GEOMETRY. 
PROBLEM    I. 


Given,  the  hypotenuse,  and  the  sum  of  the  other  two  sides 
of  a  right-angled  triangle,  to  determine  the  triangle. 

Let  ABO  be  the  A.  Put  OB  =  y,  AB 
=  x,AO  =  h,  and  CB  +  AB  =  s.  Then, 
by  a  given  condition,  we  have  \y 

x  +  y  =  s; 
and,  z2+  tf=  h\  (Th.  39,  B.  I).      A' — fr 

Reducing  these  two  equations,  and  we  have 

x  =  \s  db  Jn/2F=7";         y  -  J»  db  Jv/2A2  —  s2. 

If  A  =  5  and  s  =  7,  #  =  4  or  3,  and  y  =  3  or  4. 

Remark.  —  In  place  of  putting  a;  to  represent  one  side,  and  y  the 
other,  we  might  put  [x-\-  y)  to  represent  the  greater  side,  and  (x  —  y) 
the  less  side  ;  then, 

A2 
z2  +  y2  =  o  j  and  2x  =  s,  etc. 


PROBLEM    II. 

G-iven,  the  base  and  perpendicular  of  a  triangle,  to  find  the 
side  of  its  inscribed  square. 

Let  ABO  be  the  A.  Put 
AB  =  b,  the  base,  OB  =  p, 
the  perpendicular. 

Draw  FF  parallel  to  AB, 
and  suppose  it  equal  to  FGr,     A 
a  side  of  the  required  square ;  and  put  FF '=  x. 

Then,  by  similar  A's,  we  have 

01:  FF  : :  OB  :  AB. 

That  is,  p  —  x  :  x  : :  p  :  b. 

Hence,        bp  —  bx  =  px:  or,  x  =  _    ,     . 
^6  +  p 

That  is,  tfAe  side  of  the  inscribed  square  is  equal  to  the 

product  of  the  base  and  altitude,  divided  by  their  sum. 


BOOK  VIII 


233 


PROBLEM    I'll 


In  a  triangle,  having  given  the  sides  about  the  vertical 
angle,  and  the  line  bisecting  that  angle  and  terminating  in 
the  base,  to  find  the  base. 

Let  ABO  be  the  a,  and  let  a  cir- 
cle be  circumscribed  about  it.  Di- 
vide the  arc  AEB  into  two  equal 
parts  at  the  point  E,  and  draw  EO. 
This  line  bisects  the  vertical  angle, 
(Cor.,  Th.  9,  B.  III).     Draw  BE. 

Put  AD  =  x,  DB  =  y,  AC  =  a, 
OB  =  b,  CD  =  c,  and  BE  =  w.     The  two  A's,  ABO  and 
EBO,  are  equiangular;  from  which  we  have 

w  +  c  :  b  : :  a  :  c;  or,  cw  +  c2  =  ab ;     ( 1 ) 

But,  as  EO  and  AB  are  two  chords  that  intersect  each 
other  in  a  circle,  we  have 

cw  =  xy,     (Th.  IT,  B.  HI). 
Therefore,  xy  +  c2  =  ab.     (2) 

But,  as  OB  bisects  the  vertical  angle,  we  have 

a  :  b  : :  x  :  y,     (Th.  24,  B.  II). 


Or, 
Hence, 

And, 


ay 

y 


(3) 


P*   +    C2   =    a$;     Qr?   y   =    S^/b2__W 


=  v/i 


x 


aA*-?. 


Now,  as  a;  and  y  are  determined,  the  base  is  deter- 
mined. 

Remark.— Observe  that  equation  (2)  is  Theorem  20,  Book  III. 
20* 


234  GEO  M  E  T 11  Y. 


PROBLEM    IV. 


To  determine  a  triangle,  from  the  base,  the  line  bisecting 
the  vertical  angle,  and  the  diameter  of  the  circumscribing  circle. 

Describe  the  circle  on  the  given 
diameter,  AB,  and  divide  it  into  two 
parts,  in  the  point  D,  so  that  AD  x 
DB  shall  be  equal  to  the  square  of 
one  half  the  given  base,  (Th.  17,  B.  III). 

Through   D  draw  JEDG,   at  right  *-J- 

angles  to  AB,  and  EG  will  be  the  given  base  of  the 
triangle. 

Put        AD  =  n,  DB  =  m,  AB  =  d,  DG  =  b. 

Then,  n  +  m  =  d,  and  nm  =  b2 ; 

and  these  two  equations  will  determine  n  and  m ;  there- 
fore, we  shall  consider  n  and  m  as  known. 

Now,  suppose  HUG  to  be  the  required  A;  and  draw 
HIB  and  HA.  The  two  A's,  ABH,  DBI,  are  equian- 
gular ;  and,  therefore,  we  have 

AB  :  HB  ::  IB  :  DB. 

But  SI  is  a  given  line,  that  we  will  represent  by  c ; 
and  if  we  put  IB  —  w,  we  shall  have  HB  =  c  +  w;  then 
the  above  proportion  becomes, 

d  :  c  +  w  : :  w  :  m. 

!Now,  w  can  be  determined  by  a  quadratic  equation ; 
and,  therefore,  IB  is  a  known  line. 

In  the  right-angled  A  DBI,  the  hypotenuse  IB,  and 
the  base  DB,  are  known ;  therefore,  DI  is  known,  (Th. 
39,  B.  I) ;  and  if  i)  J  is  known,  i?J  and  IGr  are  known. 

Lastly,  let  UH=  x,  EG  =  y,  and  put  EI=p,  and  IG 

Then,  by  Theorem  20,  Book  HE,  pq  +  c2  =  xy      ( 1 ) 
But,  x  :  y  ::  p  :  q  (Th. 24,  B. II). 


BOOK  VIII.  235 

Or.  x=*l  (2) 

Now,  from  equations  ( 1 )  and  ( 2 )  we  can  determine  x 
and  y,  the  sides  of  the  A ;  and  thus  the  determination  has 
been  attained,  carefully  and  easily,  step  by  step. 


PROBLEM   V. 

Three  equal  circles  touch  each  other  ex  ernally,  and  thus 
inclose  one  acre  of  ground;  what  is  the  diameter  in  rods  of 
each  of  these  circles  f 

Draw  three  equal  circles  to  touch  each  other  exter- 
nally, and  join  the  three  centers,  thus  forming  a  triangle. 
The  lines  joining  the  centers  will  pass 
through  the  points  of  contact,  (Th.  7, 

b.  ni).  7 

Let  R  represent  the  radius  of  these         N* 
equal  circles ;  then  it  is  obvious  that        / 

each  side  of  this  A  is  equal  to  2R.      /  I A 

The  triangle  is  therefore  equilateral, 

and  it  incloses  the  given  area,  and  three  equal  sectors. 

As  the  angle  of  each  sector  is  one  third  of  two  right 
angles,  the  three  sectors  are,  together,  equal  to  a  semi- 
circle ;  but  the  area  of  a  semi-circle,  whose  radius  is  R,  is 

irR2 

expressed  by  -j— ;   and  the   area  of  the  whole  triangle 

A 

must  be         -f  160 ;  but  the  area  of  the  A  is  also  equal  to 

A 

R   multiplied  by  the  perpendicular  altitude,  which  is 
R^l. 

Therefore,      R2^S  =  ^  +  160. 

A 


Or,       JR2(2v/3  —  *)  =  320. 

g-  ,_   820 

2^3  —  3.1415926 
Hence,  R  =  31.48  4-  rods,  for  the  required  result. 


JP  =       .      32° =  ?^2-  _  992.248. 

2^3  —  3.1415926     0.3225 


236  GEOMETRY. 

Problem  VI.  —  In  a  right-angled  triangle,  having  given 
the  base  and  the  sum  of  the  perpendicular  and  hypotenuse, 
to  find  these  two  sides. 

Prob.  VII. — Given,  the  base  and  altitude  of  a  triangle,  to 
divide  it  into  three  equal  parts,  by  lines  parallel  to  the  base. 

Prob.  VIII. — In  any  equilateral  A,  given  the  length  of 
the  three  perpendiculars  drawn  from  any  point  within,  to  the 
three  sides,  to  determine  the  sides. 

Prob.  IX. — In  a  right-angled  triangle,  having  given  the 
base,  ( 3 ),  and  the  difference  between  the  hypotenuse  and  per- 
pendicular,  (1),  to  find  both  these  two  sides. 

Prob.  X.  —  In  a  right-angled  triangle,  having  given  the 
hypotenuse,  (5),  and  the  difference  between  the  base  and 
perpendicular,  ( 1 ),  to  determine  both  these  two  sides. 

Prob.  XI. — Having  given  the  area  of  a  rectangle  inscribed 
in  a  given  triangle,  to  determine  the  sides  of  the  rectangle. 

Prob.  XII. — In  a  triangle,  having  given  the  ratio  of  the 
two  sides,  together  with  both  the  segments  of  the  base,  made 
by  a  'perpendicular  from  the  vertical  angle,  to  determine  the 
sides  of  the  triangle. 

Prob.  XHL — In  a  triangle,  having  given  the  base,  the 
sum  of  the  other  two  sides,  and  the  length  of  a  line  drawn 
from  the  vertical  angle  to  the  middle  of  the  base,  to  find  the 
sides  of  the  triangle. 

Prob.  XIV. — To  determine  a  right-angled  triangle,  having 
given  the  lengths  of  two  lines  drawn  from  the  acute  angles  to 
the  middle  of  the  opposite  sides. 

Prob.  XV. — To  determine  a  right-angled  triangle,  having 
given  the  perimeter,  and  the  radius  of  the  inscribed  circle. 

Prob.  XVI. —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  ratio  of  the  two  sides. 

Prob.  XV  JUL. —  To  determine  a  right-angled  triangle,  having 
given  the  hypotenuse,  and  the  side  of  the  inscribed  square. 


BOOK  VIII.  237 

Prob.  XVIII.  —  To  determine  the  radii  of  three  equal  cir- 
cles inscribed  in  a  given  circle,  and  tangent  to  each  other,  and 
also  to  the  circumference  of  the  given  circle, 

Prob.  XIX. — In  a  right-angled  triangle,  having  given  the 
perimeter,  or  sum  of  all  the  sides,  and  the  perpendicular  let 
fall  from  the  right  angle  on  the  hypotenuse,  to  determine  the 
triangle  ;  that  is,  its  sides, 

Prob.  XX. — To  determine  a  right-angled  triangle,  having 
given  the  hypotenuse,  and  the  difference  of  two  lines  drawn 
from  the  two  acute  angles  to  the  center  of  the  inscribed  circle, 

Prob.  XXE.  —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  difference  of  the  two  other 
sides, 

Prob.  XXII.  —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  rectangle,  or  product  of  the 
two  sides, 

Prob.  XXIII. — To  determine  a  triangle,  having  given  the 
lengths  of  three  lines  drawn  from  the  three  angles  to  the  mid- 
dle of  the  opposite  sides, 

Prob.  XXIV.  —  In  a  triangle,  having  given  all  the  three 
sides,  to  find  the  radius  of  the  inscribed  circle. 

Prob.  XXV. — To  determine  a  right-angled  triangle,  having 
given  the  side  of  the  inscribed  square,  and  the  radius  of  the 
inscribed  circle. 

Prob.  XXVI.  —  To  determine  a  triangle,  and  the  radius 
of  the  inscribed  circle,  having  given  the  lengths  of  three  lines 
drawn  from  the  three  angles  to  the  center  of  that  circle, 

Prob.  XXVTI.  —  To  determine  a  right  -  angled  triangle, 
having  given  the  hypotenuse,  and  the  radius  of  the  inscribed 
circle. 

Prob.  XXVUI. — The  lengths  of  tivo  parallel  chords  on  the 
same  side  of  the  center  being  given,  and  their  distance  apart, 
to  determine  the  radius  of  the  circle. 

Prob.  XXIX.  —  The  lengths  of  two  chords  in  the  same 


238  GEOMETRY. 

circle  being  given,  and  also  the  difference  of  their  distances 
from  the  center,  to  find  the  radius  of  the  circle. 

Prob.XXX. — The  radius  of  a  circle  being  given,  and  also 
the  rectangle  of  the  segments  of  a  chord,  to  determine  the  dis- 
tance of  the  point  at  which  the  chord  is  divided,  from  the 
center. 

Prob.  XXXI. — If  each  of  the  two  equal  sides  of  an  isos- 
celes triangle  be  represented  by  a,  and  the  base  by  2b,  what 
will  be  the  value  of  the  radius  of  the  inscribed  circle  f 

.  t>       b^a*  —  b* 

Ans.  R  = 5 — . 

a  +  b 

Prob.  XXXII.  —  From  a  point  without  a  circle  whose 
diameter  is  d,  a  line  equal  to  d  is  drawn,  terminating  in  the 
concave  arc,  and  this  line  is  bisected  at  the  first  point  in  which 
it  meets  the  circumference.  What  is  the  distance  of  the  point 
without  from  the  center  of  the  circle? 

It  is  not  deemed  necessary  to  multiply  problems  in  the 
application  of  algebra  to  geometry.  The  preceding  will 
be  a  sufficient  exercise  to  give  the  student  a  clear  con- 
ception of  the  nature  of  such  problems,  and  will  serve  as 
a  guide  for  the  solution  of  others  that  may  be  proposed 
to  him,  or  that  may  be  invented  by  his  own  ingenuity. 

MISCELLANEOUS    PROPOSITIONS. 

We  shall  conclude  this  book,  and  the  subject  of  Geom- 
etry, by  offering  the  following  propositions,  —  some  the- 
orems, others  problems,  and  some  a  combination  of  both, 
— not  only  for  the  purpose  of  impressing,  by  application, 
the  geometrical  principles  which  have  now  been  estab- 
lished, but  for  the  not  less  important  purpose  of  culti- 
vating the  power  of  independent  investigation. 

After  one  or  two  propositions  in  which  the  beginner 
will  be  assisted  in  the  analysis  and  construction,  we  shall 
leave  him  to  his  own  resources,  with  the  caution  that  a 


BOOK    VIII. 


239 


—  D 


patient  consideration  of  all  the  conditions  in  each  case, 
and  not  mere  trial  operation,  is  the  only  process  by  which 
he  can  hope  to  reach  the  desired  result. 

1.  From  two  given  points,  to  draw  two  equal  straight 
lines,  which  shall  meet  in  the  same  point  in  a  given 
straight  line. 

Let  A  and  B  be  the  given  points, 
and  CD  the  given  straight  line.  Pro- 
duce the  perpendicular  to  the  straight 
line  AB  at  its  middle  point,  until  it 
meets  CD  in  G.  It  is  then  easily 
proved  that  G  is  the  point  in  CD  in 
which  the  equal  lines  from  A  and 
B  must  meet.  That  is,  that  AG 
=  BG. 

If  the  points  A  and  B  were  on 
opposite  sides  of  CD,  the  directions 
for  the  construction  would  be  the 
same,  and  we  should  have  this  fig- 
ure; but  the  reasoning  by  which 
we  prove  AG  =  BG  would  be  un- 
changed. 

2.  From  two  given  points  on  the  same  side  of  a  given 
straight  line,  to  draw  two  straight  lines  which  shall  meet 
in  the  given  line,  and  make  equal  angles  with  it. 

Let  CD  be  the  given  line,  and 
A  and  B  the  given  points. 

From  B  draw  BE  perpendicular 
to  CD,  and  produce  the  perpen- 
dicular to  F,  making  EF  equal  to 
BE)  then  draw  AF,  and  from  the 
point  G,  in  which  it  intersects 
CD,  draw  GB.  Now,  [__B  GE  = 
l_EGF=[_AGC  Hence,  the 
angles  B  GD  and  A  G  C  are  equal, 
and  the  lines  AG  and  BG  meet 
in  a  common  point  in  the  line  CD,  and  made  equal  angles  with 
that  line. 


240  GEOMETRY. 

3.  If,  from  a  point  without  a  circle,  two  straight  lines 
be  drawn  to  the  concave  part  of  the  circumference,  making 
equal  angles  with  the  line  joining  the  same  point  and  the 
center,  the  parts  of  these  lines  which  are  intercepted  within 
the  circle,  are  equal. 

4.  Ka  circle  be  described  on  the  radius  of  another  circle, 
any  straight  line  drawn  from  the  point  where  they  meet, 
to  the  outer  circumference,  is  bisected  by  the  interior  one. 

5.  From  two  given  points  on  the  same  side  of  a  line 
given  in  position,  to  draw  two  straight  lines  which  shall 
contain  a  given  angle,  and  be  terminated  in  that  line. 

6.  If,  from  any  point  without  a  circle,  lines  be  drawn 
touching  the  circle,  the  angle  contained  by  the  tangents  is 
double  the  angle  contained  by  the  line  joining  the  points 
of  contact  and  the  diameter  drawn  through  one  of  them. 

7.  If,  from  any  two  points  in  the  circumference  of  a 
circle,  there  be  drawn  two  straight  lines  to  a  point  in  a 
tangent  to  that  circle,  they  will  make  the  greatest  angle 
when  drawn  to  the  point  of  contact. 

8.  From  a  given  point  within  a  given  circle,  to  draw  a 
straight  line  which  shall  make,  with  the  circumference, 
an  angle,  less  than  any  angle  made  by  any  other  line 
drawn  from  that  point. 

9.  If  two  circles  cut  each  other,  the  greatest  line  that 
can  be  drawn  through  either  point  of  intersection,  is  that 
which  is  parallel  to  the  line  joining  their  centers. 

10.  If,  from  any  point  within  an  equilateral  triangle, 
perpendiculars  be  drawn  to  the  sides,  their  sum  is  equal 
to  a  perpendicular  drawn  from  any  of  the  angles  to  the 
opposite  side. 

11.  If  the  points  of  bisection  of  the  sides  of  a  given  tri- 
angle be  joined,  the  triangle  so  formed  will  be  one  fourth 
of  the  given  triangle. 

12.  The  difference  of  the  angles  at  the  base  of  any  tri- 
angle, is  double  the  angle  contained  by  a  line  drawn  from 
the  vertex  perpendicular  to  the  base,  and  another  bisect- 
ing the  angle  at  the  vertex. 


BOOK  VIII.  241 

13.  If,  from  the  three  angles  of  a  triangle,  lines  be 
drawn  to  the  points  of  bisection  of  the  opposite  sides, 
these  lines  intersect  each  other  in  the  same  point. 

14.  The  three  straight  lines  which  bisect  the  three 
angles  of  a  triangle,  meet  in  the  same  point. 

15.  The  two  triangles,  formed  by  drawing  straight 
lines  from  any  point  within  a  parallelogram  to  the  ex- 
tremities of  two  opposite  sides,  are,  together,  one  half  the 
parallelogram. 

16.  The  figure  formed  by  joining  the  points  of  bisection 
of  the  sides  of  a  trapezium,  is  a  parallelogram. 

17.  If  squares  be  described  on  three  sides  of  a  right- 
angled  triangle,  and  the  extremities  of  the  adjacent  sides 
be  joined,  the  triangles  so  formed  are  equal  to  the  given 
triangle,  and  to  each  other. 

18.  If  squares  be  described  on  the  hypotenuse  and  sides 
of  a  right-angled  triangle,  and  the  extremities  of  the  sides 
of  the  former,  and  the  adjacent  sides  of  the  others,  be 
joined,  the  sum  of  the  squares  of  the  lines  joining  them 
will  be  equal  to  five  times  the  square  of  the  hypotenuse. 

19.  The  vertical  angle  of  an  oblique-angled  triangle 
inscribed  in  a  circle,  is  greater  or  less  than  a  right  angle, 
by  the  angle  contained  between  the  base  and  the  diam- 
eter drawn  from  the  extremity  of  the  base. 

20.  If  the  base  of  any  triangle  be  bisected  by  the  diam- 
eter of  its  circumscribing  circle,  and,  from  the  extremity 
of  that  diameter,  a  perpendicular  be  let  fall  upon  the 
longer  side,  it  will  divide  that  side  into  segments,  one  of 
which  will  be  equal  to  one  half  the  sum,  and  the  other  to 
one  half  the  difference,  of  the  sides. 

21.  A  straight  line  drawn  from  the  vertex  of  an  equi- 
lateral triangle  inscribed  in  a  circle,  to  any  point  in  the 
opposite  circumference,  is  equal  to  the  sum  of  the  two  lines 
which  are  drawn  from  the  extremities  of  the  base  to  the 
same  point. 

22.  The  straight  line  bisecting  any  angle  of  a  triangle 
21  Q 


242  GEOMETRY. 

inscribed  in  a  given  circle,  cuts  the  circumference  in  a 
point  which  is  equi-distant  from  the  extremities  of  the 
side  opposite  to  the  bisected  angle,  and  from  the  center 
of  a  circle  inscribed  in  the  triangle. 

23.  If,  from  the  center  of  a  circle,  a  line  be  drawn  to 
any  point  in  the  chord  of  an  arc,  the  square  of  that  line, 
together  with  the  rectangle  contained  by  the  segments 
of  the  chord,  will  be  equal  to  the  square  described  on  the 
radius. 

24.  If  two  points  be  taken  in  the  diameter  of  a  circle, 
equidistant  from  the  center,  the  sum  of  the  squares  of  the 
two  lines  drawn  from  these  points  to  any  point  in  the  cir- 
cumference, will  be  always  the  same. 

25.  If,  on  the  diameter  of  a  semicircle,  two  equal  circles 
be  described,  and  in  the  space  included  by  the  three  cir- 
cumferences, a  circle  be  inscribed,  its  diameter  will  be  § 
the  diameter  of  either  of  the  equal  circles. 

26.  If  a  perpendicular  be  drawn  from  the  vertical  angle 
of  any  triangle  to  the  base,  the  difference  of  the  squares 
of  the  sides  is  equal  to  the  difference  of  the  squares  of 
the  segments  of  the  base. 

27.  The  square  described  on  the  side  of  an  equilateral 
triangle,  is  equal  to  three  times  the  square  of  the  radius 
of  the  circumscribing  circle. 

28.  The  sum  of  the  sides  of  an  isosceles  triangle  is  less 
than  the  sum  of  any  other  triangle  on  the  same  base  and 
between  the  same  parallels. 

29.  In  any  triangle,  given  one  angle,  a  side  adjacent  to 
the  given  angle,  and  the  difference  of  the  other  two  sides, 
to  construct  the  triangle. 

30.  In  any  triangle,  given  the  base,  the  sum  of  the 
other  two  sides;  and  the  angle  opposite  the  base,  to  con- 
struct the  triangle. 

31.  In  any  triangle,  given  the  base,  the  angle  opposite 
to  the  base,  and  the  difference  of  the  other  two  sides,  to 
construct  the  triangle. 


PLANE   TRIGONOMETRY 


AND 


SPHERICAL    GEOMETRY  AND   TRIGONOMETRY. 


(243) 


TKIGONOMETKY. 


PAET    I. 

PLANE    TRIGONOMETRY. 


SECTION  I. 

ELEMENTARY   PRINCIPLES. 

Trigonometry,  in  its  literal  and  restricted  sense,  has 
for  its  object  the  measurement  of  triangles.  When  it 
treats  of  plane  triangles  it  is  called  Plane  Trigonometry. 
In  a  more  enlarged  sense,  trigonometry  is  the  science 
which  investigates  the  relations  of  all  possible  arcs  of  the 
circumference  of  a  circle  to  certain  sti%ight  lines,  termed 
trigonometrical  lines  or  circular  functions,  connected  with 
and  dependent  on  such  arcs,  and  the  relations  of  these 
trigonometrical  lines  to  each  other. 

The  measure  of  an  angle  is  the  arc  of  a  circle  inter- 
cepted between  the  two  lines  which  form  the  angle — the 
center  of  the  arc  always  being  at  the  point  where  the . 
two  lines  meet. 

The  arc  is  measured  by  degrees,  minutes,  and  seconds; 
there  being  360  degrees  to  the  whole  circle,  60  minutes 
in  one  degree,  and  60  seconds  in  one  minute.  Degrees, 
minutes,  and  seconds,  are  designated  by  °,  ',  " ;  thus, 
27°  14'  21",  is  read  27  degrees  14  minutes  21  seconds. 

The  circumferences  of  all  circles  contain  the  same 
number  of  degrees,  but  the  greater  the  radius  the  greater 

(244) 


SECTION  I.  -.  245 

is  the  absolute  length  of  a  degree.  The  circumference  of 
a  carriage  wheel,  the  circumference  of  the  earth,  or  the 
still  greater  and  indefinite  circumference  of  the  heavens, 
has  the  same  number  of  degrees ;  yet  the  same  number 
of  degrees  in  each  and  every  circumference  is  the  meas- 
ure of  precisely  the  same  angle. 

DEFINITIONS. 

1.  The  Complement  of  an  arc  is  90°  minus  the  arc. 

2.  The  Supplement  of  an  arc  is  180°  minus  the  arc. 

3.  The  Sine  of  an  angle,  or  of  an  arc,  is  a  line  drawn 
from  one  end  of  an  arc,  perpendicular  to  a  diameter 
drawn  through  the  other  end.  Thus,  BF  is  the  sine  of 
the  arc  AB,  and  also  of  the  arc  BBE.  BK  is  the  sine 
of  the  arc  BB. 

4.  The  Cosine  of  an  arc  is  the  per- 
pendicular distance  from  the  center  of 
the  circle  to  the  sine  of  the  arc ;  or,  it  is 
the  same  in  magnitude  as  the  sine  of 
the  complement  of  the  arc.  Thus,  OF 
is  the  cosine  of  the  arc  AB;  but  CF= 
KB,  is  the  sine  ojLBB. 

5.  The  Tangent  of  an  arc  is  a  line  touching  the  circle 
in  one  extremity  of  the  arc,  and  continued  from  thence,  to 
meet  a  line  drawn  through  the  center  and  the  other  ex- 
tremity. Thus,  AH  is  the  tangent  to  the  arc  AB,  and 
BL  is  the  tangent  of  the  arc  BB. 

6.  The  Cotangent  of  an  arc  is  the  tangent  of  the  com- 
plement of  the,  arc.  Thus,  BL,  which  is  the  tangent  of 
the  arc  BB,  is  the  cotangent  of  the  arc  AB. 

Remark. — The  co  is  but  a  contraction  of  the  word  complement. 

7.  The  Secant  of  an  arc  is  a  line  drawn  from  the  center 
of  the  circle  to  the  extremity  of  the  tangent.  Thus,  CH 
is  the  secant  of  the  arc  AB,  or  of  its  supplement  BBE. 

8.  The  Cosecant  of  an  arc  is  the  secant  of  the  comple- 
ment.    Thus,  CLy  the  secant  of  BB,  is  the  cosecant  of  AB. 

21* 


PLANE    TRIGONOMETRY. 

9.  The  Versed  Sine  of  an  arc  is  the  distance  from  the 
extremity  of  the  arc  to  the  foot  of  the  sine.  Thus,  AF 
is  the  versed  sine  of  the  arc  AB,  and  DK  is  the  versed 
sine  of  the  arc  DB. 

For  the  sake  of  brevity,  these  technical  terms  are  con- 
tracted thus :  for  sine  AB,  we  write  sin.  AB ;  for  cosine 
AB,  we  write  cos.  AB;  for  tangent  AB,  we  write  tan. 
AB,  etc. 

From  the  preceding  definitions  we  deduce  the  follow- 
ing obvious  consequences : 

1st.  That  when  the  arc  AB  becomes  insensibly  small, 
or  zero,  its  sine,  tangent,  and  versed  sine  are  also 
nothing,  and  its  secant  and  cosine  are  each  equal  to 
radius. 

2d.  The  sine  and  versed  sine  of  a  quadrant  are  each 
equal  to  the  radius ;  its  cosine  is  zero,  and  its  secant  and 
tangent  are  infinite. 

3d.  The  chord  of  an  arc  is  twice  the  sine  of  one  half 
the  arc.     Thus,  the  chord,  BCr,  is  double  the  sine,  BF. 

4th.  The  versed  sine  is  equal  to  the  difference  between 
the  radius  and  the  cosine. 

5th.  The  sine  and  cosine  of  any  arc  form  the  two  sides 
of  a  right-angled  triangle,  which  has  a  radius  for  its 
hypotenuse.  Thus,  OF  and  FB  are  the  two  sides  of  the 
right-angled  triangle,  CFB. 

Also,  the  radius  and  tangent  always  form  the  two 
sides  of  a  right-angled  triangle,  which  has  the  secant  of 
the  arc  for  its  hypotenuse.  This  we  observe  from  the 
right-angled  triangle,  CAR. 

To  express  these  relations  analytically,  we  write 

sin.2  +  cos.2  =  B2  (1) 

El     +  tan.2  =  sec.2  (2) 

From  the  two  equiangular  triangles  CFB,  CAR,  we 

have 

OF  :  FB  =  CA  :  AH. 


SECTION   I. 

That  is, 
cos.  :  sin.  = 
Also, 

=  R  : 
OF 

tan. ; 
:  OB 

whence, 
=   OA  : 

tan. 
OR. 

.R.sin. 
cos. 

That  is, 

• 

cos.  :  R  = 

=  R  : 

:  sec. ; 

whence, 

COS. 

sec.  =  I 

247 

(3) 


The  two  equiangular  triangles,  OAR  and  ODD,  give 

OA  :  AH  =  DL  :  DO. 

That  is, 

R  :  tan.  =  cot.  :  R;  whence,  tan.  cot.  =  R\     (5) 
Also,  OF  :  FB  =  DL  :  DO. 

That  is, 
cos.  :  sin.  =  cot.  :  R;  whence,  cos.  R  =  sin.  cot.   (6) 
From  equations  (4)  and  (5),  we  have 

cos.  sec.  m  tan.  cot.  (7) 

Or,  cos.  :  tan.  =  cot.  :  sec. 

ver.  sin.  =  1  —  cos.  (8) 

The  ratios  between  the  various  trigonometrical  lines 
are  always  the  same  for  arcs  of  the  same  number  of 
degrees,  whatever  be  the  length  of  the  radius ;  and  we 
may,  therefore,  assume  radius  of  any  length  to  suit  our 
convenience.  The  preceding  equations  will  be  more  con- 
cise, and  more  readily  applied,  by  making  the  radius 
equal  unity.  This  supposition  being  made,  we  have,  for 
equations  1  to  6,  inclusive, 


sin.2  +  cos.2 

m   1. 

(1) 

1  +  tan.2 

=  sec.2 

(2) 

sin.      ,  o , 

tan.  = (3) 

cos. 

1 

cos.   =    

sec. 

(4) 

tan.  =     1       (5) 

cos.  ==  sin.  cot. 

(6) 

cot 


Let  the  circumference,  AFDH,  be  divided  into  four 
equal  parts  by  the  diameters,  AD  and  FIT,  the  one  hori- 


248 


PLANE   TRIGONOMETRY. 


w 

R 

'/? 

n 

m/ 

m, 

C 

\ 

n' 

B" 

^^ 

[3/// 

zontal  arid  the  other  vert- 
ical. These  equal  parts 
are  called  quadrants,  and 
they  may  be  distinguished 
as  the  first,  second,  third, 
and  fourth  quadrants. 

The  center  of  the  circle 
is  taken  as  the  origin  of 
distances,  or  the  zero  point, 
and  the  different  directions 
in  which  distances  are  esti- 
mated from  this  point  are  indicated  by  the  signs  +  and 
— .  If  those  from  0  to  the  right  be  marked  +,  those 
from  0  to  the  left  must  be  marked  — ;  and  if  distances 
from  O  upwards  be  considered  plus,  those  from  0  down- 
wards must  be  considered  minus. 

If  one  extremity  of  a  varying  arc  be  constantly  at  A, 
and  the  other  extremity  fall  successively  in  each  of  the 
several  quadrants,  we  may  readily  determine,  by  the 
above  rule,  the  algebraic  signs  of  the  sines  and  cosines 
of  all  arcs  from  0°  to  360°.  Now,  since  all  other  trigo- 
nometrical lines  can  be  expressed  in  terms  of  the  sine 
and  cosine,  it  follows  that  the  algebraic  signs  of  all  the 
circular  functions  result  from  those  of  the  sine  and 
cosine. 

We  shall  thus  find  for  arcs  terminating  in  the 

ein. 

1st  quadrant,  + 

2d         "  + 

3d        "  — 

4th       "  — 

PROPOSITION   I. 

The  chord  of  60°  and  the  tangent  of  45°  are  each  equal  to 
radius ;  the  sine  of  30°,  the  versed  sine  of  60°,  and  the  co- 
sine of  60°  are  each  equal  to  one  half  the  radius. 


COS. 

tan. 

cot. 

sec. 

cosec. 

vers. 

+ 

+ 

+ 

+ 

+ 

+ 

— 



— 

— 

+ 

+ 

— 

+ 

+ 



— 

+ 

+ 



— 

+ 



4- 

SECTION   I. 


249 


With  C  as  a  center,  and  CA  as  a 
radius,  describe  the  arc  ABF,  and     F 
from  A  lay  off  the  arcs  AD  =  45°, 
AB  =  60°,  and  AE  =  90° ;  then 
is  EB  =  30°. 

1st.  The  side  of  a  regular  in- 
scribed hexagon  is  the  radius  of 
the  circle,  (Prob.  28,  B.  IV),  and  as  the  arc  subtended 
by  each  side  of  the  hexagon  contains  60°,  we  have  the 
chord  of  60°  equal  to  the  radius. 

2d.  The  triangle  OAH  is  right-angled  at  A,  and  the 
angle  O  is  equal  to  45°,  being  measured  by  the  arc  AD ; 
hence  the  angle  at  H  is  also  equal  to  45°,  and  the  trian- 
gle is  isosceles.  Therefore  AH  =  CA  =  radius  of  the 
circle. 

3d.  The  triangle  ABC  is  isosceles,  and  Bn  is  a  per- 
pendicular from  the  vertex  upon  the  base ;  hence  An  = 
n  C  —  Bm,  But  Bm  is  the  sine  of  the  arc  BE,  Cn  is  the 
cosine  of  the  arc  AB,  and  An  is  the  versed  sine  of  the 
same  arc,  and  each  is  equal  to  one  half  the  radius. 
-  Hence  the  proposition ;  the  chord  of  60°,  etc. 


PROPOSITION   II. 

Given,  the  sine  and  the  cosine  of  two  arcs,  to  find  the  sine 
and '  the  cosine  of  the  sum  and  of  the  difference  of  the  sanv 
arcs  expressed  by  the  sines  and  cosines  of  the  separate  arcs. 

Let  Gr  be  the  center  of  the 
circle,  CD  the  greater  arc, 
and  DF  the  less,  and  denote 
these  arcs  by  a  and  b  re- 
spectively. 

Draw  the  radius  GrD ;  make 
the  arc  DE  equal  to  the  arc 
DF,  and  draw  the  chord  EF, 
From  F  and  E,  the  extremi- 
ties, and  J,  the  middle  point 


G    M 


NO 


250  PLANE    TRIGONOMETRY. 

of  the  chord,  let  fall  the  perpendiculars  FM,  FP,  and 
IN,  on  the  radius  GO.  Also  draw  DO,  the  sine  of  the 
arc  CD,  and  let  fall  the  perpendiculars  Iff  on  FM,  and 
FK  on  IK 

Now,  by  the  definition  of  sines  and  cosines,  DO  = 
sin.a;  GrO  =  coa.a;  FI  =  sin.6;  GI  =  cos.5.  "We  are 
to  find 

M"  =  sin.  (a  +  b);  GM  =  cos.  {a  -f  5); 
^P  =  sin.  (a  —  6) ;  GP  =  cos.  (a  —  b). 

Because  IN  is  parallel  to  DO,  the  two  a's,  GDO, 
GIN,  are  equiangular  and  similar.  Also,  the  A  FHI  is 
similar  to  the  A  GIN;  for  the  angles,  FIG  and  HIN, 
are  right  angles ;  from  these  two  equals,  taking  away  the 
common  angle  HIL,  we  have  the  angle  FIB.  =  the  angle 
GIN  The  angles  at  H  and  N  are  right  angles ;  there- 
fore, the  A's  FHI,  GIN,  and  GDO,  are  equiangular 
and  similar;  and  the  side  HI  is  homologous  to  IN 
and  DO. 

Again,  as  FI  =  IF,  and  IK  is  parallel  to  FM7 

FH=  IK,  and  HI  =  KF. 

By  similar  triangles  we  have 

GD  :  DO  =   #7  :  7ZV". 
That  is,  R  :  sin.a  =  cos.5  :  IN;  or,  7^=  sin-yggj.     ( l ) 
Also,  GD  :  GO  =  FI :  Fff 

That  is,  i£  :  cos.a  =  sin.6  :  HF;  or,  FH- 

Also,  GD  :  GO  =  GI :  GN 

That  is,  R :  cos.a  =  cos.5 :  GN;  or,  GN= 

Also,  GD  :  DO  =  FI  :  IH. 

That  is,  R  :  sin.a  =  sin.5  :  7#;  or,  75*=  8m'a  sin'h.    (4) 

By  adding  the  first  and  second  of  these  equations,  we 
have 

IN+  FH=  FM=  sin.  (a  +  b). 


That  is,  R  :  cos.a  =  sin.6  :  HF;  or,  FH=  2£^HL_5.     (2) 


That  is,  R :  coa.a  =  cos.6 :  #iV;  or,  GN=  C08,^C— 5.     ( 3 ) 


SECTION    I.  251 

mi    ,.              •     /     ,  \       sin. a  cos.5  +  cos.a  sin. b 
That  is,         sin.  (a  +  )  = ^ 

By  subtracting  the  second  from  the  first,  since 
IN—  FH=  IN—  1K=  UP,  we  have 

.     r        TN       sin.fl  cos.5  —  cos.tf  sin.5 
bib.  (a— I)  = ^ 

By  subtracting  the  fourth  from  the  third,  we  have 

G-N —  IH  =  GrM  =  cos.  (a  +  b)  for  the  first  member. 

rr                      /     ,   tn       cos.tf  cos. b  —  sin. a  sin. b      ,KX 
Hence,     cos.  (a  +  o)  = s .     (5) 

By  adding  the  third  and  fourth,  we  have 

GN+  I&=  GN+NP=GP  =  cos.(a— b). 

t-t                 /        tn      cos.a  cos.5  +  sin.a  sin.6  ;a4 

Hence,  cos. (a  —  b)  = s .         ( 6 ) 

Collecting  these  four  expressions,  and  considering  the 
radius  unity,  we  have 

{sin. (a  +  b)=- sin.a  eos.b  +  cos.a  sin.5  ( 7 ) 

sin.(a — b)—  sin.a  cos.6 — cos.a  sin.5  (8) 

cos.(a  +  b)  —  cos.a  cos.6  —  sin.a  sin. b  ( 9 ) 

cos.(a — b)  =  cos.a  cos.5  -f  sin.a  sin. b  ( 10 ) 

Formulae  (A)  accomplish  the  objects  of  the  proposi- 
tion, and  from  these  equations  many  useful  and  import- 
ant deductions  can  be  made.  The  following  are  the 
most  essential : 

By  adding  ( 7 )  to  ( 8 ),  we  have  ( 11 ) ;  subtracting  ( 8 ) 
from  ( 7 )  gives  ( 12 ).  Also,  ( 9 )  added  to  ( 10 )  gives  ( 13 ) ; 
( 9 )  taken  from  ( 10 )  gives  ( 14 ). 

r  sin.(a  +  b)  +  sin. (a — b)  —  2sin.a  cos.5        ( 11 ) 


(*) 


sin.(a  +  5)  —  sin. (a — b)  =  2cos.a  sin. b  (12) 
cos.(a  +  b)  +  cos.(a —  b)  —  2cos.a  cos.5  ( 13 ) 
cos. (a — b)  —  cos.(a  +  5)  =  2sin.a  sin.5        (14) 

If  we  put  a  +  b  =  A,  and  a  —  b  =  B,  then  ( H )  become* 
(15),  (12)  becomes  (16),  (13)  becomes  (17),  and  (14)  be- 
comes  ( 18 ). 


3  sin.^  +  sm.5=2sin.(^i-?)cos.(^— — )      (15) 


(0) 


252  PLANE    TRIGONOMETRY. 

A  +  B\  /A- 

)  COS.  ( 

2     }        >     2 
J.  -  sin.£  =  2cos.  (^t^)  sin.  (^— ?)      ( 16 ) 

.A  +  cos.^  =  2cos.  (^—^  cos.  (^— ?)     ( 17 ) 

cos.^— cos.  J.  =  2sin.  (^-^)  sin.  (^—?)      ( 18 ) 

If  we  divide  ( 15 )  by  (I6 ),  (observing  tbat  55l  =  tan., 

cos. 

cos 

and  -7— 1  =  cot.  = as  we  learn  by  equations  (6)  and 

tan.  -     u 


sm 
cos 


sin. 


( 5  )7  we  sball  bave 


sin.^1  +  sin.i? 


sm, 


rA+B 


fA—B^ 


A+B> 


fJL+lf\  (A—B\   .        /A+B\ 

(__)cos.(.-r-)tan.(-T-) 


sin.^1  — sin.i?  /A+B\     .     /A—B\~       ,A—B 


(19) 


COS. 


/A+B\~.     (A—B\   .       /A—Bx 

Whence, 
sin.A-f  sin.i?  :  sin.J.  — sin.i?  =  tan.  L )  :  tan.  (— — -) 

That  is :  The  sum  of  the  sines  of  any  two  arcs  is  to  the  dif- 
ference of  the  same  sines,  as  the  tangent  of  one  half  the  sum 
of  the  same  arcs  is  to  the  tangent  of  one  half  their  difference. 

By  operating  in  the  same  way  with  the  different  equa- 
tions in  formulae  (<7),  we  find, 

fsin.J.  + sin.i?  / A  -f  B\ 

-^tan^-^— ) 

A-B^ 

COS.B—C08.A  ™  cot'  V- 2 


(■*>) 


COS.  J.  +  COS. 

sin.A  +  sin.i?  /A  —  B\ 


sin.vl — sin.i?  /A  —  B\ 

cos. A  +  cosT5  "■ tan*  V     2~  /. 
sin.JL — sin.i?  ,A  +  B\ 

co^B^co^A  "  cot  \     2/ 

cos.  J.  -f  cos.ff  __  cot*  \ 2/ 
cos.i? — cos. A  "~  /A-^B 


tan. 


*.—$ 


) 


(20) 
(21) 
(22) 
(23) 

(24) 


SECTION   I.  253 

These  equations  are  all  true,  whatever  be  the  value 
of  the  arcs  designated  by  A  and  B ;  we  may,  therefore, 
assign  any  possible  value  to  either  of  them,  and  if  in 
equations  (20)?  (21),  and  (24),  we  make  B  =  0,  we  shall 
have, 

8in^  tan.^  =  — L,  (25) 


1  +  cos.^.  2      cot.  \A 

sin. A             ,  A          1  ,OA* 

cot.  -^  = — r  ( 26 ) 


@  1 — cos.  A  2      tan.  J  J. 

1  +  cos. A  __  cot. \A 


(27) 


1 — cos.  A      tan.  J  J.      tan2.  \  A 
If  we  now  turn  back  to  formulae  (A),  and  divide  equa- 
tion ( 7 )  by  ( 9 ),  and  ( 8 )  by  ( 10 ),  observing  at  the  same 

sin 
time  that  — -  =  tan.,  we  shall  have, 
cos.  ' 

tan.(a  +  5)  =  Sin^  cos'5  +  cos^  sin*5 


tan.(# — 5)  = 


cos. a  cos. b  —  sin.a  sin. b 
sin.#  cos.5 —  cos.a  sin.5 


cos.a  cos.5 -fsin.a  sin. b 
By  dividing  the  numerators  and  denominators  of  the 
second  members  of  these  equations  by  (cos.a  cos.5),  we 
find, 

sin.a  cos.5    cos.a  sin.5 

,    .  ,x    cos.a  cos.5    cos.a  cos.5      tan.a-ftan.5 

tan.(a+5)= , — i *ri""i — r       +      i     (28) 

v        J    cos.a  cos.5    sin. a  sin. 5    1 — tan.atan.5 

cos.a  cos.5    cos.a  cos.5 
sin.a  cos.5    cos.a  sin.5 

,      ,N    cos.a  cos.5    cos.a  cos.5      tan.  a— tan.  5 

tan.(a— 5)= 1 , . — ,=r-— -—. ,     (29 ) 

x        '    cos.a  cos.5    sm.fl  sm.5    1-ftan.a  tan.5 

cos.a  cos.5    cos.a  cos.5 

If  in  equation  ( 11 ),  formulae  ( B ),  we  make  a  =  5,  we 
shall  have, 

sin. 2a  =  2sin.a  cos.a         (30) 

Making  the  same  hypothesis  in  equation  ( 13 ),  gives, 

cos.2a  +  1=  2c<3s2.a         (31) 

22 


254  PLANE   TRIGONOMETRY. 

The  same  hypothesis  reduces  equation  (14)  to 

1  — cos.2a  =  2sin2.a         (32) 

The  same  hypothesis  reduces  equation  ( 28 )  to 

,      0  2tan.#  ,oo\ 

tan.2a  =  - — - — -—  ( 33 ) 

1  —  tan\a 

If  we  substitute  a  for  2a  in  ( 31 )  and  ( 32 )y  we  shall  have 

1  +  cos.a  =  2cos.2Ja.         (34) 

and  1  —  cos.a  =  2sin.2Ja.         ( 35 ) 

PROPOSITION   III. 

In  any  right-angled  plane  triangle,  we  may  have  the  fol- 
lowing proportions : 

1st.  The  hypotenuse  is  to  either  side,  as  the  radius  is  to  the 
sine  of  the  angle  opposite  to  that  side. 

2d.  One  side  is  to  the  other  side,  as  the  radius  is  to  the  tan- 
gent of  the  angle  adjacent  to  the  first  side. 

3d.  One  side  is  to  the  hypotenuse,  as  the  radius  is  to  the 
secant  of  the  angle  adjacent  to  that  side. 

Let  CAB  represent  any  right- 
angled  triangle,  right-angled  at 
A. 

(Here,  and  in  all  cases  hereafter,  we  shall  represent  the  angles  of  a 
triangle  by  the  large  letters  A,  B,  C,  and  the  sides  opposite  to  them, 
by  the  small  letters  a,  b,  c.) 

From  either  acute  angle,  as  0,  take  any  distance,  as 
CD,  greater  or  less  than  CB,  and  describe  the  arc  BF. 
This  arc  measures  the  angle  C.  From  B,  draw  BF  par- 
allel to  BA ;  and  from  JE,  draw  EG,  also  parallel  to  BA 
oyBF. 

By  the  definitions  of'  sines,  tangents,  secants,  etc,  BF 
is  the  sine  of  the  angle  C;  EG  is  the  tangent,  CG  the 
secant,  and  OF  the  cosine. 


SECTION    I. 


255 


Now,  by  proportional  triangles  we  have, 

OB:BA=  OD:DF  or,  a  :  c  =  R  :  sin.C 
OA:AB=OF:FG  or,  b  :  c  =  i2  :  tan.  0 
OA:  OB  =  OB:  OG     or,  b  :  a  =  R  :  sec.<7 

Hence  the  proposition. 

Scholium. — If  the  hypotenuse  of  a  triangle  is  made  radius,  one  side 
is  the  sine  of  the  angle  opposite  to  it,  and  the  other  side  is  the  cosine 
of  the  same  angle.     This  is  obvious  from  the  triangle  CDF, 


PROPOSITION    IV. 

In  any  triangle,  the  sines  of  the  angles  are  to  one  another 
as  the  sides  opposite  to  them. 

Let  ABO  be  any  tri- 
angle. From  the  points 
A  and  B,  as  centers, 
with  any  radius,  de- 
scribe the  arcs  meas- 
uri  ng  these  angles,  and  j* 
draw  pa,  OD,  and  mn, 
perpendicular  to  AB. 

Then,  pa  =  sin.JL,  and  mn  =  sin.i?. 

By  the  similar  A's,  Apa  and  A  OB,  we  have, 

R  :  sin. J.  =  b  :  OD ;  or,  R{OD)  =  b  sin.J.     (1) 

By  the  similar  a's,  Bmn  and  BOD,  we  have, 

R  :  sin.B  =  a:  OD;  or,  R(OD)  =  a  sin.B     (2) 

By  equating  the  second  members  of  equations  ( 1 ) 
and  (2) 

b  sin. A  =  a  sin.i?. 

Hence,  sin.A  :  sin.i?  —  a  :b 

Or,  a  :  b  =  biii.A  :  sin.i?. 

Scholium  1. — When  either  angle  is  90°,  its  sine  is  radius. 

Scholium  2. — When  CB  is  less  than  AC,  and  the  angle  B,  acute, 
the  triangle  is  represented  by  A  CB.  When  the  angle  B  becomes  B', 
it  is  obtuse,  and  the  triangle  is  ACB/ ;  but  the  proportion  is  equally 


256  PLANE    TRIGONOMETRY. 

true  with  either  triangle  ;  for  the  angle  CB'D  =  CBA,  and  the  sine 
of  CB'D  is  the  same  as  the  sine  of  AB/  C.  In  practice  we  can  deter- 
mine which  of  these  triangles  is  proposed,  by  the  side  AB  being 
greater  or  less  than  AC;  or,  by  the  angle  at  the  vertex  C  being  large, 
as  A  CB,  or  small,  as  A  CB'. 

In  the  solitary  case  in  which  AC,  CB,  and  the  angle  A,  are  given, 
and  CB  less  than  AC,  we  can  determine  both  of  the  A's  ACB  and 
A  CB/ ;  and  then  we  surely  have  the  right  one. 


PROPOSITION    V. 

If  from  any  angle  of  a  triangle,  a  perpendicular  he  let  fall 
on  the  opposite  side,  or  base,  the  tangents  of  the  segments  of 
the  angle  are  to  each  other  as  the  segments  of  the  base. 

Let  ABC  be  the  triangle.  Let  fall 
the  perpendicular  CD,  on  the  side 
AB. 

Take  any  radius,  as  Cn,  and  de- 
scribe the  arc  which  measures  the  A  QX 
angle  C.  From  n,  draw  qnp  parallel  to  AB.  Then  it  is 
obvious  that  np  is  the  tangent  of  the  angle  DCB,  and  nq 
is  the  tangent  of  the  angle  ACB. 

Now,  by  reason  of  the  parallels  AB  and  qp,  we  have, 
qn  :  np  =  AB  :  BB 

That  is,         tan.^OZ)  :  tsm.BCB  =  AB  :  BB. 

PROPOSITION   VI. 

If  a  perpendicular  be  let  fall  from  any  angle  of  a  triangle 
to  its  opposite  side  or  base,  this  base  is  to  the  sum  of  the  other 
two  sides,  as  the  difference  of  the  sides  is  to  the  difference  of 
the  segments  of  the  base. 

(See  figure  to  Proposition  5.) 

Let  AB  be  the  base,  and  from  C,  as  a  center,  with  the 
shorter  side  as  radius,  describe  the  circle,  cutting  AB  in 
(x,  and  AC  in  F;  produce  AC  to  E. 


SECTION    I.  257 

It  is  obvious  that  AE  is  the  sum  of  the  sides  AG  and 
OB,  and  AF  is  their  difference. 

Also,  AD  is  one  segment  of  the  base  made  by  the  per- 
pendicular, and  BB  =  BGr  is  the  other;  therefore,  the 
difference  of  the  segments  is  AG. 

As  A  is  a  point  without  a  circle,  by  Cor.  Th.  18,  B. 
Ill,  we  have 

AE  x  AF  =  AB  x  AG 

Hence,  ^LB    :   AE  -  AF   :    J.#. 

PROPOSITION    VII. 

2%£  mm  0/  awy  free  szefes  0/  a  triangle  is  to  their  difference, 
as  the  tangent  of  one  half  the  sum  of  the  angles  opposite  to 
these  sides,  is  to  the  tangent  of  one  half  their  difference. 

Let   ABO   be    any   plane    triangle.  ^ 

Then,  by  Proposition  4,  we  have, 

BO:  A 0  =  sin.J.  :  sin.^. 
Hence,  A~  ~~B 

BC  + A  C:  BC— A  0=  sin.A+sm.B :  sin.^— -sin.^  (Th.  9,B.  II). 
But, 

tan.  ( — ^ —  j  :  tan.  ( — - —  )  ==  sin. A  +  sin.i?  :  sin.  J. 

—  sin.B,  (eq.  (19),  Trig.) 

Comparing  the  two  latter  proportions,  (Th.  6,  B.  H), 
we  have, 

BO+AO:BO—  AO=t&n.  (f-j^)  :  tan.  (^-^) 

Hence  the  proposition. 

PROPOSITION   VIII. 

Given,  the  three  sides  of  any  plane  triangle,  to  find  some 
relation  which  they  must  bear  to  the  sines  and  cosines  of  the 
respective  angles. 

22*  r 


258 


PLANE    TRIGONOMETRY. 


Let  ABO  be 
the  triangle,  and 
let  the  perpen- 
dicular fall  either 
upon,  or  without 
the  base,  as  shown  c 
in    the     figures. 


C    «    b    x 

By  recurring  to  Th.  40,  B.  I,  we  shall  find 

a2  -f  b2  —  c* 


OB  = 


2a 


(1) 


£fow,  by  Proposition  3,  we  have 

R  :  cos.  C  =  b  :  CD. 


Therefore, 


OB  = 


b  cos.  0 
~~R 


(2) 


Equating  these  two  values  of  OB,  and  reducing,  we 
have 


2ab 


(m) 


In  this  expression  we  observe,  that  the  part  c,  whose 
square  is  found  in  the  numerator  with  the  minus  sign,  is 
the  side  opposite  to  the  angle ;  and  that  the  denominator 
is  twice  the  rectangle  of  the  sides  adjacent  to  the  angle. 
From  these  observations  we  at  once  draw  the  following 
expressions  for  the  cosine  A,  and  cosine  B : 

A         R(b2  +  c2  —  a?) 
cos.  J.  =  -v 


cos.  B 


2bc 
R(a2  +  c* 


b2) 


2ac 


(n) 


(P) 


As  these  expressions  are  not  convenient  for  logarith- 
mic computation,  we  modify  them  as  follows : 
If  we  put  2a  =  JL,  in  equation  ( 31 ),  we  have 


cos.  A  +  1  =  2cos.2  %  A. 


In  the  preceding  expression,  (»),  if  we  consider  radius 
unity,  and  add  1  to  both  members,  we  shall  have 


SECTION   I.  259 

COS.  ^+1    =    1+   _JT_ 

Therefore,      2cos.'  \A  =  S^fcHl^ 

26<? 

' U   (b  +  c)2  —  a2 
2bc 
Considering  b  -f  c  as  one  quantity,  and  observing  that 
(6  +  c)2 — a2  is  the  difference  of  two  squares,  we  have 

(6+c)2— a2=(&+c+a)  (&+c— a) ;  but  (&+*— «)= &+c+a— 2a. 

Hence,      2cos.^  -  <5  +  c  +  «><* L+  °  +  "~H 
2  2fo 

Or,  cos.2 1 j.  -  : ? Li I r 

By  putting — -  m  s,  and  extracting  square  root, 

A 

the  final  result  for  radius  unity  is 

cos.  \A  =  \1WE& 
v         bo 

For  any  other  radius  we  must  write 


cos.  |,i=  y/ifo»-«), 

By  inference,   cos.JJ?  =  \j ?L^Z^1. 


Also,  cos.  J(7  i  v/^^^l). 

In  every  triangle,  the  sum  of  the  three  angles  is  equal 
to  180° ;  and  if  one  of  the  angles  is  small,  the  other 
two  must  be  comparatively  large ;  if  two  of  them  are 
small,  the  third  one  must  be  large.  The  greater  angle 
is  always  opposite  the  greater  side ;  hence,  by  merely 
inspecting  the  given  sides,  any  person  can  decide  at 
once  which  is  the  greater  angle ;  and  of  the  three  pre- 


260  PLANE    TRIGONOMETRY. 

ceding  equations,  that  one  should  be  taken  which  applies 
to  the  greater  angle,  whether  that  be  the  particular 
angle  required  or  not ;  because  the  equations  bring  out 
the  cosines  to  the  angles ;  and  the  cosines  to  very  small 
arcs  vary  so  slowly,  that  it  may  be  impossible  to  decide, 
with  sufficient  numerical  accuracy,  to  what  particular 
arc  the  cosine  belongs.  For  instance,  the  cosine  9.999999, 
carried  to  the  table,  applies  to  several  arcs ;  and,  of 
course,  we  should  not  know  which  one  to  take ;  but  this 
difficulty  does  not  exist  when  the  angle  is  large ;  there- 
fore, compute  the  largest  angle  first,  and  then  compute 
the  other  angles  by  Proposition  4. 

But  we  can  deduce  an  expression  for  the  sine  of  any 
of  the  angles,  as  well  as  the  cosine.  It  is  done  as  fol- 
lows: 

EQUATIONS  FOR  THE   SINES  OF  THE  ANGLES. 

Resuming  equation  ( m ),  and  considering  radius  unity, 
we  have 

cos.  0  =  __ 

Subtracting  each  member  of  this  equation  from  unity, 
gives 

Make  2a  =  O,  in  equation  (32) ;  then  a  =  \Qy 
and    1  —  cos.  0  =  2sin.2|<7.  (2) 

Equating  the  second  members  of  (1)  and  (2), 
2ab  —  a?  —  b2  +  e* 


2sin.2i<7  = 


2ab 

<?2  —  (a  —  bf 
2ab 

(c  -f  b  —  a)  (c  +  a  —  b) 
2ab  I' 


SECTION   I.  261 

/c  -\-  b  —  a\  /c  +  cl  —  b\ 
Or,  sin,  10  =  {£7Y-){—S—)m 

ab 

t>„+  e+l> — a      c+b+a              iC+a — b       c+a+b      , 
But.__=  _ o.and—g—  =  — ^ h. 

Put =  s,  as  before;  then, 

sin.i<7=  JBS 
v  ab 

By  taking  equation  (p ),  and  proceeding  in  the  same 

manner,  we  have 

sin.ji?=  \J¥E^VEh. 

v  ao 

From  t4  sin.  JJ.  =   \JEEM3, 

v  c6 

The  preceding  results  are  for  radius  unity ;  for  any 
other  radius,  we  must  multiply  by  the  number  of  units 
in  such  radius.  For  the  radius  of  the  tables  we  write 
H;  and  if  we  put  it  under  the  radical  sign,  we  must 
write  B2;  hence,  for  the  sines  corresponding  with  our 
logarithmic  table,  we  must  write  the  equations  thus, 


v  be 

v  ac 

sm.  10  =  J '#<»=ME3. 

v  ab 

A  large  angle  should  not  be  determined  by  these 
equations,  for  the  same  reason  that  a  small  angle  should 
not  be  determined  from  an  equation  expressing  the 
cosine. 

In  practice,  the  equations  for  cosine  are  more  gener- 
ally used,  because  more  easily  applied. 


262  PLANE   TRIGONOMETRY. 

The  formulae  which  we  have  thus  analytically  devel- 
oped, express  nearly  all  the  important  relations  between 
the  sines,  cosines,  and  tangents  of  arcs  or  angles ;  and 
we  have  also  demonstrated  all  the  theorems  required  for 
the  determination  of  the  unknown  parts  of  any  plane 
triangle,  three  of  the  parts  of  which  are  given,  one  at 
least  being  a  side. 

Such  relations  might  be  indefinitely  multiplied,  but 
those  already  established  are  sufficient  for  most  practical 
purposes,  and  when  others  are  required,  no  difficulty 
will  be  found  in  deducing  them  from  these. 

The  following  geometrical  demonstrations  of  many  of 
the  preceding  relations,  are  offered,  in  the  belief  that 
they  will  prove  useful  disciplinary  exercises  to  the  stu- 
dent. 

1st.  Let  the  arc  AD=A;  then  2)  #= sin.  J.;  CG-=cos.A; 
J9J=sin.}J.;^2)=2sin.JJL;  OZ=cos.|J.; 
CI=DO;  and  Z>£=2DO=2cos.}A 
The  angle,  DBA,  is  measured  by 
one  half  the  arc  AD ;  that  is,  by  \A. 
Also,        AD  a  =  DBA  =  J  A. 
Now,  in  the  triangle,  BD  Gr,  we  have 

sm.DBG  :  2>#=sin.90°  :  BD. 
That  is,  sin.  J  A :  sin.  J.=l :  2cos.  \A. 
Or,  sin.J=2sin.JJ.  cos.JJ.; 

which  corresponds  to  equation  ( 30 ). 

In  the  same  triangle, 
sin.90°  :  BD=sin.BDG  :  BG;  and  sin.BDG=co8.DBG. 
That  is,  1  :  2cos.|J=eos.|J.  :  1+cos.JL 
Or,  2cos.2  %A=l+co8.Ay  same  as  equation  (34). 

In  the  triangle,  DGrA,  we  have, 

sin.90°  :  AD  =  sin.  GDA  :  a  A. 
That  is,  1  :  2sin. JJ.  =  sin.JJ.  :  1— cos. J.. 
Or,  2sin.2  \A  =  1 — cos.^4,  same  as  equation  ( 35 ). 


SECTION   I. 


263 


2  :  2sin. \A  =  2sin.JJ.  :  versed  sin. A. 
versed  sin. A  =  2sin.2  \A. 


By  similar  triangles,  we  have, 

BA  :  AD  =  AD:  AG. 

That  is, 

Or, 

2d.  From  O  as  the  center,  with  OA  as  the  radius, 
describe  a  circle.  Take  any  arc, 
AB,  and  call  it  A  ;  and  AD  a  less 
arc,  and  call  it  B ;  then  BD  is  the 
difference  of  the  two  arcs,  and  must 
be  designated  by  (A — B) ;  arc  AG 
=  arc  AB ;  therefore, 

arcD#  =  A  +  B;  FG  =  sm.A; 

En  ss=  sin. B ;  Gn  =  sin. J.  -f  sin. B ; 
Bn  =  sin.  J.  —  sin.i?. 
Fm  =  mD  =  OJI=  cos.B ;  mn  =  cos.  J. ; 
therefore,   Fm  +  mn=  cos.  J.  -f  cos.i?  =  Fn ; 
mD  —  mn  =  eos.B  —  cos.  A  =  nD ; 
A  +  B> 


and 

Because, 
therefore, 

-or, 


I>#  =  2sin. 


pm 


JVF=AD;  AB  +  NF=A  +  B; 

180°  —  (J.  +  i?)  =  arc  ZE; 

90°-(£±*)~larcJ®. 


But  the  chord,  FB,  is  twice  the  sine  of  J  arc  _Fi? ; 

that  is,      FB  =  2sin.  (90°  —  A  ±  ffi)  =  2cos.  (^-^). 

The  L_7i6rD  =  LjSZZ),  because  both  are  measured 
by  one  half  of  the  arc  BD;  that  is,  by  (  ~  \  and  the 
two  triangles,  6rftD  and  .Fwi?,  are  similar. 

The  angle,  GFn,  is  measured  by  f — - — \ 

In  the  triangle,  FBG,  Fn  m  drawn  from  an  angle  per- 

OF  THE        ^A 


264  PLANE    TRIGONOMETRY. 

pendicular  to  the  opposite  side ;  therefore,  by  Proposition 
5,  we  have, 

Gn  :  nB  =  tan.  GFn  :  t&n.BFn. 

That  is,  sin.-4.-f  sin.i? :  sin. A — sin.JB=tan.( — i — )  :  tan. 

I — - — ).     This  is  equation  (19 ). 

In  the  triangle,  GnB,  we  have, 

sin.90°  :  BG  =  am.nBG  :  Gn;  &m.nDG=cos.nGD. 

That  is,  1  :  2sin.  (^~)  =  cos.  (A~B)  :  sin.^l-f  sin.J?. 

Or,  sin.  J.  -f  sin.2?  =  2sin.  ( — - — )  cos.  ( — - — ), 

the  same  as  equation  (15). 

3d.  In  the  triangle,  FnB,  we  have, 

sin.90  :  FB  =  sin.^Frc  :  Bn. 

That  is,  1  :  2cos.(^±^)  =  sin.(4ip?)  -  sin. A—  sin.#. 

Or,  sin.JL-sin.£  =  2cos.  (^~~)  sin.  (A~B), 

the  same  as  equation  (16). 

4th.  In  the  triangle,  FBn,  we  have, 

sin.90  :  FB  —  cos.BFn  :  Fn. 

That  is,  1  :  2cos.  (^jp  )  =  cos.(^2?)  :  cos.^L+cos.5. 

Or,  cos.  J.  +  cos.5  =  2cos.  ( — - — \  cos.  (     ~    V  the 
same  as  equation  ( 17 ). 

5th.  In  the  triangle,  GnB,  we  have, 

sin.90°  :  GD  =  mi.nGB  :  nB. 

That  is,  1  :  2sin.  (_ — )  =  sin.  (— ~^—  \  :  cos.B—cos.A, 
the  same  as  equation  (18). 

6th.  In  the  triangle,  FGn,  we  have, 

sin.  GFn  :  Gn  =  cos.  GFn  :  Fn. 


SECTION    I.  265 

That  is,  sin.  — ^—  :  sin.  A+sin.B  —  cos.  — —  :  cos.A+ 
2  2i 

COS.B. 

Or,  (sin. J.  -f  ain.B)  cos.  (—J — )  =  (cos. A  -f  cos.i?)  sift. 

.     A  +  B 

Or,    sin'f  ±  si^|  =  _*       -  tan.  (*±*\     the 
cos.^  +  cos..B  J.  +  .S  V      2      /' 

cos. — o — 

same  as  equation  (20). 

7th.  In  the  triangle,  FnB,  we  have, 

Fn  :  nB  ::  1  :  tan.BFn. 

That  is,  cos.J5+cos.JL  :  sin. A — sin.i?  ::  1  :  tan.|(A — B), 

r.  sin.  A  —  sin.i?      ,        /A — B\     .* 

Or,  = -  =  tan.  ( — - — ),   the    same 

cos.  A  +  cos._#  V     2     /' 

as  equation  (22). 

8th.  In  the  triangle,  GrnD,  we  have, 

Gn  :  nB  : :  1  :  tan.w6rD. 

That  is, 
sin.  J. -f  sin. B  :  cos. B  —  cos.  J.  ::  1  :  tanY — - — V 

cos.  B  —  cos.  A        ,       /A  —  B\ 
sin.  A  +  sin.  B  =  ^  (~J~  ).' 

NATURAL    SINES,    COSINES,    ETC. 

When  the  radius  of  the  circle  is  taken  as  the  unit  of 
measure,  the  numerical  values  of  the  trigonometrical 
lines  belonging  to  the  different  arcs  of  the  quadrant,  be- 
come natural  sines,  cosines,  etc.  They  are  then,  in  fact, 
but  numbers  expressing  the  number  of  times  that  these 
lines  contain  the  radius  of  the  circle  in  which  they  are 
taken.  The  tables  usually  contain  only  the  sines  and 
cosines,  because  these  are  generally  sufficient  for  practi- 
23 


266  PLANE    TRIGONOMETRY. 

cal  purposes,  and  the  others,  when  required,  are  readily 
expressed  in  terms  of  them. 

We  proceed  to  explain  a  method  for  computing  a  table 
of  natural  sines  and  cosines. 

It  was  shown,  in  Book  V,  that  the  linear  value  of  the 
arc  180°,  in  a  circle  whose  radius  is  unity,  is 

3.141592653. 

This  divided  by  180  x  60,  the  number  of  minutes  in 
180°,  will  give  the  length  of  one  minute  of  arc,  which  is 

.00029088820867. 

But  there  can  be  no  sensible  difference  between  the 
length  of  the  arc  V  and  its  sine ;  and,  within  narrow 
limits,  that  sine  will  increase  directly  with  the  arc. 

Hence, 


sin. 

V 

= 

.0002908882. 

sin. 

2' 

= 

.0005817764. 

sin. 

3' 

= 

.0008726646. 

sin. 

4' 

= 

.0011635528. 

sin. 

5' 

= 

.0014544410. 

sin. 

6' 

= 

.0017453292. 

sin. 

V 

= 

.0020362175. 

sin. 

8' 

= 

.0023271057. 

sin. 

9' 

= 

.0026179938. 

sin. 

10' 

= 

.0029088811. 

Beyond  this,  the  error  which  would  arise  from  taking 
the  arc  for  its  sine,  upon  which  the  above  proceeds, 
would  affect  the  final  decimal  figures;  and  we  must, 
therefore,  continue  the  computation  of  the  series  by 
other  processes.  To  find  the  values  of  the  cosines  of 
arcs,  from  V  to  10',  we  have 


cos.  =  *S\  —  sin.2  =  1  —  J  sin.2,  nearly. 

That  is,  when  the  sines  are  very  small  fractions,  as  is 
the  case  for  all  arcs  below  10',  we  can  find  the  cosine  by 
subtracting  one  half  of  the  square  of  the  sine  from  unity. 


SECTION   I.  267 

Whence,        cos.  1'  =  .9999999577. 

cos.  2'  =  .9999998308. 

cos.  3'  -  .9999993204. 

cos.  4'  =  .99999932304. 

cos.  5'  =  .99999894290. 

cos.  6'  =  .99999847753. 

cos.  V  =  .99999792735. 

cos.  8'  -  .9999973035. 

cos.  9'  =  .9999965730. 

cos.  10'  =  .9999957703. 

The  natural  sines  of  arcs,  differing  by  1',  from  10'  up 
to  1°,  may  be  computed  from  those  of  arcs  less  than 
10',  by  means  of  equation  ( 11 ),  group  B,  which  is 

sin.  (a  -f  b)  =  2sin.  a  cos.  b  —  sin.  (a  -f  b) ; 

And  when  a  =  5,  this  equation  becomes 

sin.  2a  =  2sin.  a  cos.  b.     Eq.  ( 30 ). 

To  find  the  sine  of  11',  we  make  a  =  6',  and  5  =  5'; 

then  sin.  11'  =  2sin.  6'  cos.  5'—  sin.  1'=  .00319976913. 

0=6  =  6',       sin.  12'  =  2sin.  6'  cos.  6'. 
a  =  7',  b  =  6',  sin.  13'  =  2sin.  T  cos.  6'  —  sin.  1'. 
a  =  b  =  7,       sin.  14'  =  2sin.  7'  cos.  7'. 
a  =  8,  b  =  7,    sin.  15'  =  2sin.  8;  cos.  7'  —  sin.  V 

And  so  on  to  the 

sin.  30'  =  2sin.l5'cos.l5'. 
sin.l°  =  sin.  60'  =  2sin.30'cos.30'. 
sin.  2°   =  2sin.  1°  cos.  1°. 
sin.  3°   =  2sin.  2°  cos.  1°  —  sin.  1°,  etc.,  etc.,  etc. 

This  process  may  be  continued  until  we  have  found 
the  sines  and  cosines  of  all  arcs  differing  by  V,  from  0 
to  90°,  the  values  of  the  cosines  being  deduced  success- 
ively from  those  of  the  sines  by  means  of  the  formula, 


cos.  =  */l  —  sin.2. 
In  this  calculation,  we  began  by  assuming  that,  for 
small  arcs,  the  sines  and  the  arcs  were  sensibly  equal. 


268  PLANE    TRIGONOMETRY. 

It  must  be  remembered  that  this  is  but  an  approxima- 
tion ;  and  although  the  error  in  the  early  stages  of  the 
process  is  not  sufficient  to  affect  any  of  the  decimal  fig- 
ures which  enter  the  tables,  it  will  finally  become  so, 
since  it  is  constantly  increased  in  the  operations  by 
which  the  sines  and  cosines  of  the  larger  arcs  are  de- 
duced from  those  of  the  smaller.  "When  the  error  has 
been  thus  increased  until  it  reaches  the  order  of  the  last 
decimal  unit  of  the  table,  which  assigns  our  limit  of 
error,  we  must  have  the  means  of  detecting  and  correct- 
ing it. 

•  This  consists  in  calculating  the  sines  and  cosines  of 
certain  arcs  by  independent  processes,  and  comparing 
them  with  those  found  by  the  above  method. 

"We  have  seen,  for  example,  (Prop.  7,  B.  V),  that  the 
chord  of 

30°  =  .517638090;  whence,  sin.  15°  =  .258819045. 

15°  =  .2610523842;       "       "       7°  15'       =.130526192. 
7°  15'  =  .1308062583;      "       "      3°   7'  30"=  .0654031291. 

And  so  on  to 

sin.  14'  3"  45'"  =  .004090604. 

etc.  etc.  etc. 

The  following  elegant  method  of  deducing,  from  the 
sine  of  an  arc,  the  sine  and  cosine  of  one  half  the  arc,  is 
given,  assuming  that  the  student  is  familiar  with  the 
simple  algebraic  principles  upon  which  it  depends. 

Let  us  take  the  natural  sine  of  18°,  which  is  .3090170, 

18° 
and  make  x  =  sine,  and  y  the  cosine  of  9°  =  — . 

A 

Then,  x2  +  y2  =  1;  (1) 

and  2xy  =  .3090170     (2);   Eq.  (30). 

Adding,  we  have 

z2-f  2xy  +  f  =  1.3090170; 


SECTION   I.  269 

Taking  tlie  square  root,  we  have 

x  +  y  =  1.144123.     (3) 
Subtracting  ( 2 )  from  ( 1 ), 

x2  —  2xy  +  y2  =  .690983; 
taking  the  square  root, 

x  —  y  -  —.831254*         (4) 
Adding  (3)  and  (4),  2z  =    .312869, 

hence,  a;  =  sin.9°  =    .1564345 

Subtracting  (4)  from  (3),       2y  =  1.97537T 
hence,  y  =  cos.9°   =    .9876885 

Now,  by  making  #  =  the  sine  of  4°  30',  and  y  =  cosine 
of  4°  30',  and  as  before 

x2  +  y2  =  1 
and  2xy  =  .1564345, 

we  obtain  the  sine  and  cosine  of  4°  30' ;  and  another  ope- 
ration will  give  the  sine  and  cosine  2°  15',  etc.,  etc. 

We  may  in  this  manner  compute  the  sines  and  cosines 
of  all  arcs  resulting  from  the  division  of  18°  by  2,  and 
we  may  make  their  values  accurate  to  any  assigned  deci- 
mal figure. 

This  has  been  carried  far  enough  to  show  how  a  table 
of  natural  sines,  etc.,  could  be  computed ;  but  in  conse- 
quence of  the  tedious  numerical  operations  which  the 
process  requires,  other  methods  are  resorted  to  in  the 
actual  construction  of  the  table. 

The  Calculus  furnishes  formulae  giving  the  values  of 
the  sines  and  cosines  of  arcs  developed  into  rapidly  con- 
verging series,  and  from  these  the  sines  and  cosines  of 
all  arcs  from  0°  to  90°,  can  be  determined  with  great 

*  When  an  arc  is  less  than  45°,  the  cosine  exceeds  the  sine ;  and 
■when  the  arc  is  between  45°  and  90°,  the  sine   exceeds  the  cosine. 
Hence,  when  the  arc  is  9°,  y,  its  cosine,  exceeds  x,  its  sine ;  and  we 
therefore  placed  the  minus  sign  before  the  second  member  of  Eq.  (4). 
23* 


270 


PLANE    TRIGONOMETRY. 


accuracy  and  with  comparatively  little  labor.  In  the  last 
two  columns  on  each  page  of  Table  II,  will  be  found  the 
values  thus  computed  of  the  sines  and  cosines  of  every 
degree  and  minute  of  a  quadrant. 

TRIGONOMETRICAL  LINES  FOR  ARCS  EXCEEDING  90°. 


X// 


From  the  annexed  figure, 
the  construction  of  which 
needs  no  explanation,  are 
deduced  by  simple  inspec- 
tion the  results  given  in  the 
following 


TABLE 


90°  -f  a° 

270°  —  a° 

sin.  =   cos.  a,  cos.  =  —  sin.  a 

sin.  =  —  cos.  a,  cos.  =  —  sm.  a 

tan.  =  —  cot.  a,  cot.  =  —  tan.  a 

tan.  =   cot.  a,  cot.  =   tan.  a 

sec.  =  —  cosec.  a,  cosec.  =  sec.  a 

sec.  —  — cosec.  a,  cosec.  =  —  sec.  a 

180°  —  a° 

270°  +  a° 

sm.  =   sin.  a,  cos.  =  —  cos.  a 

sin.  =  —  cos.  a,  cos.  =   sin.  a 

tan.  =  —  tan.  a,  cot.  =  —  cot.  a 

tan.  =  —  cot.  a,  cot.  =  —  tan.  a 

sec.  =  —  sec.  a,  cosec.  =  cosec.  a 

sec.  =   cosec.  a,  cosec.  =  —  sec.  a 

180°  +  a0 

360°  —  a° 

sin.  =  —  sin.  a,  cos.  =  —  cos.  a 

sin.  =  —  sin.  a,  cos.  =   cos.  a 

tan.=  tan. a,  cot.  =  cot.  a 

tan.  =  —  tan.  a,  cot.  =  —  cot.  a 

sec.  = — sec.  a,  cosec.=— cosec. a 

sec.  =   sec.  a,  cosec.  =  — cosec.  a 

By  means  of  this  table,  the  values  of  the  trigonomet- 
rical lines  of  any  arc  between  90°  and  360°,  can  be  ex- 
pressed by  those  of  arcs  less  than  90°. 

If,  for  example,  the  arc  is  118°,  we  have 


SECTION    I.  271 

sin.  118°  =  sin.  (90°  +  28°)  =       cos.28°  ; 
tan.H8°  =  tan.(90°  +  28°)  =  —  cot.28° ; 
etc.,  etc.,  etc. 

For  the  arc  230°,  we  have 

sin.  230°  =  sin.  (270°  —  40°)  =  —  cos.    40°  ; 

sec.230°  =  sec.(270°  —  40°)  =  —  cosec.40°; 
etc.,  etc.,  etc. 

In  many  investigations,  it  becomes  necessary  to  con- 
sider the  functions  of  arcs  greater  than  360° ;  but  since 
the  addition  of  360°  any  number  of  times  to  the  arc  a, 
will  give  an  arc  terminating  in  the  extremity  of  a,  it  is 
obvious  that  the  arc  resulting  from  such  addition  will 
have  the  same  functions  as  the  arc  a.  And  hence  it  fol- 
lows that  the  functions  of  arcs,  however  great,  may  be 
expressed  in  terms  of  the  functions  of  arcs  less  than  90°. 


272 


PLANE    TKIGONOMETEY. 


SECTION   II. 


PLANE   TRIGONOMETRY,   PRACTICALLY  APPLIED. 


In  the  preceding  section,  the  theory  of  Trigonometry 
has  been  quite  fully  developed,  and  the  student  should 
now  be  prepared  for  its  various  applications,  were  he 
acquainted  with  logarithms.  But  logarithms  are  no  part 
of  Trigonometry,  and  serve  only  to  facilitate  the  numeri- 
cal operations.  Trigonometrical  computations  can  be 
made  without  logarithms,  and  were  so  made  long  before 
the  theory  of  logarithms  was  understood. 

For  this  reason,  we  proceed  at  once  to  the  solution  of 
the  following  triangles. 

1.  The  hypotenuse  of  a  right-angled  triangle  is  21, 
and  the  base  is  17 ;  required  the  perpendicular  and  the 
acute  angles. 

Let  CAB  be  the  triangle,  in 
which  CB  -  21,  and  CA  = 
17.  With  C  as  a  center,  and 
CD  =  1  as  a  radius,  describe 
the  arc  DE,  of  which  the  sine 
is  DFj  the  tangent  is  EG,  and 
the  cosine  is  OF, 

By  similar  triangles  we  have 
CB  :   CA 

that  is,  21  :  17 


CD 
1 


CF; 

cos.  C. 


Hence, 


17 


cos.  C  =  --  =  .80952+. 


SECTION   II.  273 

We  must  now  turn  to  Table  II,  and  find  in  the  last  two  columns 
the  cosine  nearest  to  .80952,  and  the  corresponding  degrees  and 
minutes  will  be  the  value  of  the  angle  C. 

On  page  56,  of  Tables,  near  the  bottom  of  the  page,  and  in  the 
column  with  cosine  at  the  top,  we  find  .80953,  which  corresponds 
to  35°  56'  for  the  angle  C.     The  angle  B  is,  therefore,  54°  4'. 

This  Table  is  so  arranged,  that  the  sum  of  the  degrees  at  the  top 
and  bottom  of  the  page,  added  to  the  sum  of  the  minutes  which  are 
found  on  the  same  horizontal  line  in  the  two  side  columns  of  the 
page,  make  90°. 

Thus,  in  finding  the  angle  (7,  the  number  .80953  was  found  in 
the  column  with  cosine  at  its  foot.  We  therefore  took  the  degrees 
from  the  bottom  of  the  page,  and  the  minutes  were  taken  from  the 
right  hand  column,  counting  upwards. 

For  the  side  AB}  we  have  the  proportion 

CF  :  FD  ::   CA  :  AB; 

or,  cos.  C  :  sin.  C  : :  17  :  AB; 

that  is,  .80953  :  .58708  : :  17  :  AB. 

From  which  we  find  AB  =  .58708  X  17  -J-  .80953; 

whence,  AB  =  12.328. 

If  we  had  formed  a  table  of  natural  tangents,  as  well  as  of  natu- 
ral sines,  AB  could  have  been  found  by  the  following  proportion  • 
CE  :  EG  : :   CA  :  AB 

or,  1  :  tan.  C  : :  17  :  AB) 

whence,  AB  =  17  tan.  C. 

The  perpendicular  AB  may  also  be  found  by  the  proportion 
CD  :  DF  n  CB  :  AB) 

or,  1  :  sin.  C  : :  21  :  AB; 

whence,    AB  =  21  sin.  C  =  21  x  .58708  =  12.32868. 

2.  The  two  sides  of  a  right-angled  triangle  are  150  pnd 
125 ;  required  the  hypotenuse  and  the  acute  angles. 

Let    CAB  be   the  triangle, 
which  is  the  same  as  in  the  pre-   . 
ceding  problem. 

Then,  from  the  similar  trian- 
gles, CFD  and  CAB}  we  get 

CF  :  FD  ::  CA  i  AB) 

0 


274  PLANE    TRIGONOMETRY. 

that  is,        cos.  G  :  sin.  G  : :  150  :  125  : :  6  :  5, 
which  gives  6  sin.  (7=5  cos.  C; 

hence,  36  sin.2(7  =  25  cos.*<7. 

Adding  member  to  member,     86  cos.26r  =  36  cos.2  C. 

we  have  36  (sin.2C  +  cos.2C)  ==  61  cos.*  G. 

But  sin.2  C-f  cos.2<?  =  1,  (Eq.  (1)  Trigonometry)  5 
whence,  61  cos.2C  =  36; 

cos.2  C7  =  |r  =  .5901639344; 

and  cos.  C  =  .76816,  nearly. 

Tor  find  the  angle  of  which  this  is  the  cosine,  we  turn  to  page  60 
of  tables,  and  looking  in  the  column  having  cosine  at  the  head,  we 
see  that  .76816  falls  between  .76868,  which  has  48'  opposite  to  it 
in  the  left  hand  column,  and  .76810,  which  has  49'  opposite  to  it 
in  the  same  column.  Now,  the  cosines  of  arcs  less  than  90°  de- 
crease when  the  arcs  increase,  and  the  converse ;  and  while  the 
increase  of  the  arc  is  confined  within  the  limits  of  1',  the  increase 
of  the  arc  will  be  sensibly  proportional  to  the  decrease  of  the  cosine. 
0.76828  .76828 

Hence,  0.76810  .76816 

~18       :         ~~12  : :  60"  :  x" 
which  gives  x"  =  40". 

The  angle  G  is,  therefore,  equal  to  39°  48'  40",  and  the  angle 
B  =  90°  —  39°  48'  40"  -  50°  11'  20". 
To  find  GB,  we  have 

GF  :   GD  : :   GA  :   GB 
or,  cos.  G  :  1  : :  150  :   GB 

that  is,  .78816  :  1  : :  150  :   GB 

150 
whence,  GB  =  -^^  =  195.27 +. 

3.  The  base  of  a  right-angled  triangle  is  150,  and  the 
angle  opposite  the  base  is  50°  11'  20" ;  required  the 
hypotenuse  and  the  perpendicular. 


SECTION    II. 


275 


Let  CAB  be  the  triangle. 
Then,  (Prop.  4,  Sec.  I), 
Bin.  50°  11'  20"  :  sin.  90°  ::  150  :  CB. 
Whence, 

CB  "  76816  =195-27' 
the  same  as  in  the  preceding  example. 
To  find  AB,  we  have 

CD  :  DF  ::   CB  :  ^LB; 
that  is,  1  :  sin.  C  or  cos.  B  ::  195.27  :  .4.5; 

from  which  we  find 

AB  =  195.27  sin.  39°  48'  40"; 
or,  AB  =  125.01077. 

4.  Two  sides,  the  one  30  and  the  other  35,  and  the  in- 
eluded  angle  20°,  of  a  triangle,  are  given,  to  find  the 
other  two  angles  and  the  third  side. 

Let  B A  C  be  the  triangle,  in  which  B  C 
=  35,  BA  =  30,  and  the  angle  B  = 
'20°.  From  A,  the  extremity  of  the 
shorter  side,  let  fall  on  BC  the  perpen- 
dicular AD,  thus  dividing  the  triangle 
into  the  two  right-angled  triangles  BAD  and  CAD. 

Then,  from  the  triangle  BAD,  we  have 

1st,  sin.D  :  sin.  B     ::  BA  :  AD; 

or,  1  :  sin.  20°  : :  30     :  AD  =  30  sin.  20°. 

2d,  1  :  cos.  B     : :  BA  :  BD; 

or,  1  :  cos.  20°  : :  30     :  BD  =  30  cos.  B. 

In  the  table  of  natural  sines,  we  find  sin.  20°  =  .34202,  and  the 
cos.  20°  =  .93969;  hence,  AD  =  30  X  .34202  =  10.26060,  and 
BD  =  30  x  .93969  =  28.19070,  and  therefore  DC  =  BC  — 
BD  =  6.8093. 

From  the  triangle  CAD,  we  have 

1st,  AC=  ^Alf  +^F=  \/(10.26)2  +  (6.94-)2=  12.367. 
2d,  AC  :  AD  ::  sin. 90°  :  sin.  (7; 


f6 

PLANE    TKIGONOMETRY 

or, 

12.367  :  10.26+  : :  1  :  sin.  C; 

whence, 

10  26 
sin-  °  =  12.367  r  -82968- 

and  the 

angle   G  =  56°  3'. 

If,  now,  we  add  angles  B  and  C,  and  take  the  sum  from  180°, 
the  remainder  will  be  the  angle  BA  C. 

Hence,     [_  B  AC  =1$0°  —  (56°  3'  +  20°)  =  103°  57'. 

5.  Two  sides,  the  one  18  and  the  other  24,  and  the 
angle  opposite  the  side  24  equal  to  76°,  are  given,  to  find 
the  remaining  side  and  the  other  two  angles. 
Let  x  denote  the  angle  opposite  the  side  18.     Then, 

24  :  18  : :  sin.  76°  :  sin.  xy  (Prop.  4,  Trig.), 
or,  4:3::  sin.  76°  :  sin.  x. 

sin.  a;  =  f  sin.  76°  =  f  X  .97030  =  .72772; 

whence  the  angle  opposite  the  side  18  is  46°  41'  45". 
Adding  this  to  the  given  angle,  and  taking  the  sum  from  180°, 
we  get  57°  17'  15"  for  the  third  angle. 

To  find  the  remaining  side,  denoted  by  y,  we  have 

sin.  76°  :  sin.  57°  18'  15"  ::  24  :  y; 

or,  97030  :  .84154  : :  24  :  y. 

24  x  .84154 


y  = 


.97030 


=  20.815  =  3d  side. 


6.  The  three  sides  of  a  triangle  are  18,  24,  and  20.815 ; 
required  the  angles. 

This  problem  may  be  solved  by  Prop.  6,  or  by  Prop.  8,  Trigo- 
nometry. 

First.     By  Prop.  6. 

In  the  triangle  ABC,  make  CB  = 
24,  AC  =  20.815,  and  AB  =  18.  A 

Then, 

24  :  38.815  : :  2.815  :  CD  —  BD. 
OD-BD-™*™-  4.5527. 


SECTION    II.  277 

But  CD  +  BD  =   CB  =  24. 

By  addition,  we  get       2  CD  =  28.5527; 

dividing  by  2,  and  <7Z>  =  14.2763+. 

And  hence,  .£D  =   CB  —  CD  =  24  —  14.2763  =  9.7237. 

In  the  triangle  ADB,  we  have 

BA  :     £Z>    ::  1  :  cos. 5 

or,  18  :  9.7237  : :  1  :  cos.  B  =  .54020 

rr  u    tt  x>       kq     f  cos.  57°  18'  =  .54024) 
Table  H,  Page  53,   {  cos<  5?o  1Q,  =  .54000  } 

diff.  ^  24  :  60"  : :  4  :  10" 

hence,  \__B  =  57°  18' 10". 

It  will  be  observed  that  Examples  5  and  6  refer  to  the  same  tri- 
angle, and  that  in  Example  5  the  angle  B  was  57°  18'  15".  This 
slight  discrepancy  in  the  results  should  be  expected,  on  account  of 
the  small  number  of  decimal  places  used  in  the  computations. 

Second.     By  Prop.  8. 

Sum  of  the  sides,      ,                          =62.815, 
half  sum  denoted  by  S,                      =  31.4075 
a  =  24 

S—a  =    7.4075 


Formula,  cos.  £  A  =  \  /  — — - -,  radius  being  unity. 

S(S—a)  =  31.4075  x  7.4075  =  232.65105625 

be  =  20.815    X         18  =  374.67 
£(ff  —  a)  ^    620Q5  very  near]^ 

V 152095  =  .78800. 

Hence,  cos.  \A  =  .78800,  and  \A  (Table  II,  page  59)  =  38° 
very  nearly ;  the  angle  A  is  therefore  equal  to  76°,  which  agrees 
with  Example  5. 

7.  Given,  the  three  sides,  1425, 1338,  and  493,  of  a  tri- 
angle ;  required,  the  angle  opposite  the  greater  side,  using 
the  formula  for  the  sine  of  one  half  an  angle. 
24 


278  PLANE    TRIGONOMETRY. 

Make  a  =  1425,  b  =  1338,  and  c  =  493 ;  then  the  [__  A  is 
opposite  the  side  a,  and  the  formula  is 

oc 
in  which  s  denotes  the  half  sum  of  the  three  sides. 

Then  we  have  s  ==  1628,  s  —  b  =  290,  s  —  c  =  1135,  (s  —  6) 

0_.  c)  =  329150,  6c  =  659634,  (g  —  ft)  (*  — c)  =  .498988. 


Hence,  sin.  JJ.  =  v/,498988  =  .70632. 

In  the  table  we  find         sin.  44°  56'  12"  =  .70632. 

Therefore,  \A  =  44°  56' 12",  and  A  =  89°  52'  24";— but  little 
less  than  a  right  angle. 

In  these  seven  examples  we  have  shown  that  it  is  possi- 
ble to  solve  any  plane  triangle,  in  which  three  parts,  one 
at  least  being  a  side,  are  given,  without  the  aid  of  loga- 
rithms. But,  when  great  accuracy  is  required,  and  the 
number  of  decimal  places  employed  is  large,  the  necessary 
multiplications  and  divisions,  the  raising  to  powers,  and 
the  extraction  of  roots,  become  very  tedious.  All  of  these 
operations  may  be  performed  without  impairing  the  cor- 
rectness of  results,  and  with  a  great  saving  of  labor,  by 
means  of  logarithms ;  but,  before  using  them,  the  student 
should  be  made  acquainted  with  their  nature  and  pro- 
perties. 

LOGARITHMS. 

Logarithms  are  the  exponents  of  the  powers  to  which 
a  fixed  number,  called  the  base,  must  be  raised,  to  pro- 
duce other  numbers. 

The  exponent  of  a  number  is  also  a  number  express- 
ing how  many  times  the  first  number  is  taken  as  a  factor. 

Thus,  let  a  denote  any  number ;  then  a3  indicates  that  a 
has  been  used  three  times  as  a  factor,  a4  that  it  has  been 
used  four  times  as  a  factor,  and  an  that  it  has  been  thus 
used  n  times. 


SECTION    II.  279 

Now,  instead  of  calling  these  numbers  3,  4, w, 

exponents,  we  call  them  the  logarithms  of  the  powers  a% 
a\ an. 

To  multiply  a 2  by  a 5,  we  have  simply  to  write  a,  giving 
it  an  exponent  equal  to  2  +  5 ;  thus,  a2  X  a5  =  a\ 

Hence,  the  sum  of  the  logarithms  of  any  number  of  factors 
is  equal  to  the  logarithm  of  the  product. 

To  divide  a12  by  a9,  we  have  only  to  write  a,  giving  it 
an  exponent  equal  to  12  —  9 ;  thus,  a12  -*-  a9  =  a3 ;  and, 
generally,  the  quotient  arising  from  the  division  of  a  m  by 
a%  is  equal  to  am~n. 

Hence,  the  logarithm  of  a  quotient  is  the  logarithm  of  the 
dividend  diminished  by  the  logarithm  of  the  divisor. 

If  it  is  required  to  raise  a  number  denoted  by  a3,  to  the 
fifth  power,  we  write  a,  giving  it  an  exponent  equal  to 
3x5;  thus,  (a3)5=a15,  and,  generally,  (an)m=anm. 

Hence,  the  logarithm  of  the  power  of  a  number  is  equal  to 
the  logarithm  of  the  number  multiplied  by  the  exponent  of  the 
power.  v 

To  extract  the  5th  root  of  the  number  as,  we  write  «, 
giving  it  an  exponent  equal  to  f ;  thus,  ^/a~s=a^  and, 
generally,  to  extract  any  root  of  a  number,  we  divide  the 
exponent  of  the  number  by  the  index  of  the  root,  and  the 
quotient  will  be  the  exponent  of  the  required  root. 

Hence,  the  logarithm  of  a  root  of  a  number  is  equal  to  the 
quotient  obtained  by  dividing  the  logarithm  of  the  number  by 
the  index  of  the  root 

Now,  understanding  that  by  means  of  a  table  of  loga- 
rithms we  may  find. the  numbers  answering  to  given 
logarithms,  with  as  much  facility  as  we  can  find  the  loga- 
rithms of  given  numbers,  we  see  from  what  precedes  that 
multiplications,  divisions,  raising  to  powers,  and  the  ex- 
traction of  roots,  may  be  performed  by  logarithms ;  and 
the  utility  of  logarithms,  in  trigonometrical  computations, 
mainly  consists  in  the  simplicity  and  abridgment  of  these 
operations  as  executed  by  them. 


280  PLANE    TRIGONOMETRY. 

The  common  logarithms  are  those  of  which  10  is  the 
base ;  that  is,  they  are  the  exponents  of  10. 


Thus,  lO1^  10 

Hence  the  logarithm 

L  10            =  1. 

102  =  100 

U 

u 

u 

100      =  2. 

103  =  1000 

u 

<■ 

a 

1000    =3. 

10*  =  10000 

a 

it 

a 

10000  =  4. 

etc.      etc. 

etc. 

etc.     etc. 

Since  Jq  =  1  =  101 

-'  =  10°, 

and 

am 
generally  —  =  a0  = 

1, 

it  follows  that  in  this,  as  in  all  other  systems,  the  loga- 
rithm or  1  =  0. 

From  what  precedes,  it  is  evident  that  the  logarithm  of 
any  number  between  10  and  100  must  be  found  between 

1  and  2 ;  that  is,  its  logarithm  is  1  plus  a  number  less 
than  1;  and  any  number  between  100  and  1000,  will 
have  for  its  logarithm  2  plus  some  number  less  than  1, 
and  so  on.  The  fractional  part  of  the  logarithms  of 
numbers  are  expressed  decimally. 

The  entire  number  belonging  to  a  logarithm  is  called 
its  index.  The  index  is  never  put  in  the  tables,  (except 
from  1  to  100),  and  need  not  be  put  there,  because  we 
always  know  what  it  is.  It  is  always  one  less  than  the 
number  of  digits  in  the  whole  number.  Thus,  the  num- 
ber 3754  has  3  for  the  index  to  its  logarithm,  because  the 
number  consists  of  4  digits ;  that  is,  the  logarithm  is  3  and 
some  decimal. 

The  number  347.921  has  2  for  the  index  of  its  loga- 
rithm, because  the  number  is  between  347  and  348,  and 

2  is  the  index  for  the  logarithms  of  all  numbers  over  100, 
and  less  than  1000. 

All  numbers  consisting  of  the  same  figures,  whether 
integral,  fractional,  or  mixed,  have  logarithms  consisting 
of  the  same  decimal  part.     The  logarithms  would  differ 
only  in  their  indices. 
24* 


SECTION   II.  281 

Thus,  the  number  7956.  has  3.900695  for  its  log. 
the  number  795.6  has  2.900695 
the  number  79.56  has  1.900695 
the  number  7.956  has  0.900695 
the  number  .7956  has  —1.900695 
the  number  .07956  has  —2.900695 

From  this  we  perceive  that  we  must  take  the  logarithm 
out  of  the  table  for  a  mixed  number  or  a  decimal,  the 
same  as  if  the  figures  expressed  an  entire  number;  and 
then,  to  prefix  the  index,  we  must  consider  the  value  of 
the  number. 

The  decimal  part  of  a  logarithm  is  always  positive; 
but  the  index  becomes  negative  when  the  number  is  a 
decimal;  and  the  smaller  the  decimal,  the  greater  the 
negative  index.     Hence, 

To  prefix  the  index  to  a  decimal,  count  the  decimal 
point  as  1,  and  every  cipher  as  1,  up  to  the  first  significant 
figure,  and  this  is  the  negative  index. 

For  example,  find  the  logarithm  of  the  decimal 
.0000831. 

Num.  .0000831;  log.  —5.919601. 

The  point  is  counted  one,  and  each  of  the  ciphers  is 
counted  one ;  therefore  the  index  is  minus  five. 

•The  smaller  the  decimal,  the  greater  the  negative 
index ;  and  when  the  number  becomes  0,  the  logarithm  is 
negatively  infinite. 

Hence,  the  logarithmic  sine  of  0°  is  negatively  infinite, 
however  great  the  radius. 

A  number  being  given,  to  find  its  corresponding  logarithm. 

The  logarithm  of  any  number  consisting  of  four  figures, 
or  less,  is  taken  out  of  the  table  directly,  and  without  the 
least  difficulty. 

Thus,  to  find  the  logarithm  of  the  number  3725,  we 
24* 


282  PLANE    TRIGONOMETRY. 

find  372  at  the  side  of  the  table,  and  in  the  column 
marked  5  at  the  top,  and  opposite  372,  we  find  .571126, 
for  the  decimal  part  of  the  logarithm. 

Hence,  the  logarithm  of  3725  is  3.571126. 
the  logarithm  of  37250  is  4.571126. 
the  logarithm  of  37.25   is  1.571126,  etc. 

Find  the  logarithm  of  the  number  834785. 
This  number  is  so  large  that  we  cannot  find  it  in  the 
table,  but  we  can  find  the  numbers  8347  and  8348.  The 
logarithms  of  these  numbers  are  the  same  as  the  loga- 
rithms of  the  numbers  834700  and  834800,  except  the 
indices. 

834700    log.    5.921530 
834800    log.    5.921582 

Difference,  100  52 

Now,  our  proposed  number,  834785,  is  between  the 
two  assumed  numbers ;  and,  of  course,  its  logarithm  lies 
between  the  logarithms  of  the  two  assumed  numbers; 
and,  without  further  comment,  we  may  proportion  it 
thus, 

100  :  85  =  52  :  44.2 

Or,  1.  :  .85  =  52  :  44.2 

Hence,  for  finding  from  the  table  the  logarithm  of  a 
number  consisting  of  more  than  four  places  of  figures, 
we  have  the  following 

RULE. 

Take  from  the  table  the  log.  of  the  number  expressed  by  the 
the  four  superior  figures  ;  this,  with  the  proper  index,  is  the 
approximate  logarithm.  Multiply  the  number  expressed  by  the 
remaining  figures  of  the  number,  regarded  as  a  decimal,  by 
the  tabular  difference,  and  the  product  will  be  the  correction 
to  be  added  to  the  approximate  log.  to  obtain  the  true  log. 


SECTION    II.  283 


EXAMPLES. 


1.  What  is  the  log.  of  35T.32514? 

The  log.  of  357.3  is  2.553033 

No.  not  included,     .2514 
Tabular  diff.,  122 

Prod.,  30.6708;  correction,    31 


log.  sought,  2.553064 

The  log.  of  35732.514  is  4.553064 

"  .035732514"         —2.553064. 

2.  What  is  the  log.  of  7912532  ? 

Approximate  log.,  6.898286 
.532  x  55  =   correction,  29 


True  log.  =  6.898315. 
A  logarithm  being  given,  to  find  its  corresponding  number. 

For  example,  what  number  corresponds  to  the  log. 

6.898315  ? 

The  index  6  shows  that  the  entire  part  of  the  number  must  con- 
tain seven  places  of  figures.  With  the  decimal  part,  .898315,  of 
the  log.,  we  turn  to  the  table,  and  find  the  next  less  decimal  part 
to  be  .898286,  which  corresponds  to  the  superior  places,  7912. 

The  difference  between  the  given  log.  and  the  one  next  less  is 
29.  This  we  divide  by  the  tabular  difference,  55,  because  we  are 
working  the  converse  of  the  preceding  problem.     Thus, 

29  -f-  55  =  52727+. 

Place  the  quotient  to  the  right  of  the  four  figures  before  found, 
and  we  shall  have  7912527.27  for  the  number  sought. 

This  example  was  taken  from  the  preceding  case,  and 
the  number  found  should  have  been  7912532 ;  and  so  it 
would  have  been,  had  we  used  the  true  difference,  29.26, 
in  place  of  29. 

When  the  numbers  are  large,  as  in  this  example,  the 


284  PLANE    TRIGONOMETRY. 

result  is  liable  to  a  small  error,  to  avoid  which  the  loga- 
rithms should  contain  a  great  number  of  decimal  places ; 
but  the  logarithms  in  our  table  contain  a  sufficient  num- 
ber of  decimal  places  for  most  practical  purposes. 

Hence,  for  finding  the  number  corresponding  to  any 
given  logarithm,  we  have  the  following 

RULE. 

Look  in  the  table  for  the  decimal  part  of  the  given  loga- 
rithm, and  if  not  found,  take  the  decimal  next  less,  and  take 
out  the  four  corresponding  figures. 

Take  the  difference  between  the  given  log.  and  the  next  less 
in  the  table  ;  divide  that  difference  by  the  tabular  difference, 
and  write  the  quotient  on  the  right  of  the  four  superior  fig- 
ures, and  the  result  is  the  number  sought. 

Point  off  the  whole  number  required  by  the  given  index. 

EXAMPLES. 

1.  Given,  the  logarithm  3.743210,  to  find  its  corres- 
ponding number  true  to  three  places  of  decimals. 

Ans.  5536.182. 

2.  Given,  the  logarithm  2.633356,  to  find  its  corres- 
ponding number  true  to  two  places  of  decimals. 

Ans.  429.89. 

3.  Given,  the  logarithm  —  3.291746,  to  find  its  corres- 
ponding number.  Ans.  .0019577. 

4.  What  number  corresponds  to  the  log.  3.233568  ? 

Ans.  1712.25. 

5.  What  is  the  number  of  which  1.532708  is  the  log.  ? 

Ans.  34.0963. 

6.  Find  the  number  whose  log.  is  1.067889. 

Ans.  11.692. 

EXPLANATION    OP    TABLE    II. 

Table  I  is  merely  a  table  of  numbers  and  their  corres- 
ponding logarithms,  and  requires  no  explanation  other 


SECTION   II.  285 

than  that  which  has  been  given  in  connection  with  the 
subject  of  logarithms. 

Table  II,  with  the  exception  of  the  last  two  columns, 
which  contain  natural  sines  and  cosines,  is  a  table  in 
which  are  arranged  the  logarithms  of  the  numerical 
values  of  the  several  trigonometrical  lines  corresponding 
to  the  different  angles  in  a  quadrant.  The  values  of 
these  lines  are  computed  to  the  radius  10,000,000,000, 
and  their  logarithms  are  nothing  more  than  the  loga- 
rithms, each  increased  by  10,  of  the  natural  sines,  co- 
sines, and  tangents,  of  the  same  angles;  because  the 
values  of  these  lines,  for  arcs  of  the  same  number  of  de- 
grees taken  in  different  circles,  are  directly  proportional 
to  the  radii  of  the  circles. 

The  natural  sines  are  made  to  the  radius  of  unity; 
and,  of  course,  any  particular  sine  is  a  decimal  fraction, 
expressed  by  natural  numbers.  The  logarithm  of  any 
natural  sine,  with  its  index  increased  by  10,  will  give 
the  logarithmic  sine.  Thus,  the  natural  sine  of  8°  is 
.052336. 

The  logarithm  of  this  decimal  is  —  2.718800 

To  which  add  10. 


The  logarithmic  sine  of  3°  is,  therefore,  8.718800 

In  this  manner  we  may  find  the  logarithmic  sine  of 
any  other  arc,  when  we  have  the  natural  sine  of  the 
same  arc. 

If  the  natural  sines  and  logarithmic  sines  were  on  the 
same  radius,  the  logarithm  of  the  natural  sine  would  be 
the  logarithmic  sine,  at  once,  without  any  increase  of 
the  index. 

The  radius  for  the  logarithmic  sines  is  arbitrarily 
taken  so  large  that  the  index  of  its  logarithm  is  10.  It 
might  have  been  more  or  less ;  but,  by  common  consent, 
it  is  settled  at  this  value ;  so  that  the  sines  of  the  smallest 
arcs  ever  used  shall  not  have  a  negative  index. 


286  PLANE    TRIGONOMETRY. 

In  our  preceding  equations,  sin.  a,  cos.  a,  etc.,  refei 
to  natural  sines;  and  by  such  equations  we  determine 
their  values  in  natural  numbers ;  and  these  numbers  are 
put  in  Table  II,  under  the  heads  of  nat.  sine  and  nat.  co- 
sine, as  before  observed. 

When  we  have  the  sines  and  cosines  of  an  arc,  the 
tangent  and  cotangent  are  found  by  Eq.  ( 3 ) ;  that  is, 

,  R  sin.      /A;  R  cos. 

tan.  =  (6)  cot.  =  — : : 

cos.  sm. 

and  the  secant  is  found  by  equation  (4);  that  is, 

R2 

sec.  = 

cos. 

For. example,  the  logarithmic  sine  of  6°  is  9.019235, 
and  its  cosine  9.997614.  From  these  it  is  required  to 
find  the  logarithmic  tangent,  cotangent,  and  secant. 

R  sin.  19.019235 

Cos.  subtract    9.997614 


Tan.  is  9.021621 

R  cos.  19.997614 

Sin.  subtract    9.019235 


Cotan.  is  10.978379 

R*  is  20.000000 

Cos.  subtract    9.997674 


Secant  is  10.002326 

The  secants  and  cosecants  of  arcs  are  not  given  in 
our  table,  because  they  are  very  little  used  in  practice ; 
and  if  any  particular  secant  is  required,  it  can  be  deter- 
mined by  subtracting  the  cosine  from  20 ;  and  the  cose- 
cant can  be  found  by  subtracting  the  sine  from  20. 

The  sine  of  every  degree  and  minute  of  the  quadrant 
is  given,  directly,  in  the  table,  commencing  at  0°,  and 
extending  to  45°,  at  the  head  of  the  table ;  and  from  45° 
to  90°,  at  the  bottom  of  the  table,  increasing  backward. 


SECTION    II.  287 

The  same  column  that  is  marked  sine,  at  the  top,  is 
marked  cosine  at  the  bottom ;  and  the  reason  for  this  is 
apparent  to  any  one  who  has  examined  the  definitions 
of  sines. 

The  difference  of  two  consecutive  logarithms  is  given, 
corresponding  to  ten  seconds.  Removing  the  decimal 
point  one  figure,  will  give  the  difference  for  one  second ; 
and  if  we  multiply  this  difference  by  any  proposed  num- 
ber of  seconds,  we  shall  have  a  difference  corresponding 
to  that  number  of  seconds,  above  the  logarithm  corres- 
ponding to  the  preceding  degree  and  minute. 

For  example,  find  the  sine  of  19°  17'  22". 

The  sine  of  19°  17',  taken  directly  from  the  table,  is      9.518829 
The  difference  for  10"  is  60.2 ;  for  1",  is  6.02 ;  and      • 

6.02  X  22  =  133 


Hence,  19°  17'  22"  sine  is  9.518962 

From  this  it  will  be  perceived  that  there  is  no  difficulty 
in  obtaining  the  sine  or  tangent,  cosine  or  cotangent,  of 
any  angle  greater  than  30'. 

Conversely :  Given,  the  logarithmic  sine  9.982412,  to 
find  its  corresponding  arc.  The  sine  next  less  in  the 
table  is  9.982404,  which  gives  the  arc  73°  48'.  The  differ- 
ence between  this  and  the  given  sine  is  8,  and  the  dif- 
ference for  1"  is  .61 ;  therefore,  the  number  of  seconds 
corresponding  to  8,  must  be  discovered  by  dividing  8  by 
the  decimal  .61,  which  gives  13.  Hence,  the  arc  sought 
is  73°  48'  13". 

These  operations  are  too  obvious  to  require  a  rule. 
When  the  arc  is  very  small, — and  such  arcs  as  are  sometimes 
required  in  Astronomy, — it  is  necessary  to  be  very  accu- 
rate ;  for  this  reason  we  omitted  the  difference  for  seconds 
for  all  arcs  under  30'.  Assuming  that  the  sines  and  tan- 
gents of  arcs  under  30'  vary  in  the  same  proportion  as 
the  arcs  themselves,  we  can  find  the  sine  or  tangent  of 
any  very  small  arc,  with  great  exactness,  as  follows : 


288  PLANE    TRIGONOMETRY. 

The  sine  of  V,  as  expressed  in  the  table,  is  6.463726 

Divide  this  by  60;  that  is,  subtract  logarithm  1.778151 

The  logarithmic  sine  of  1",  therefore,  is  4.685575 

Now,  for  the  sine  of  17",  add  the  logarithm  of  17     1.230449 

Logarithmic  sine  of  17",  is  5.916024 

In  the  same  manner  we  may  find  the  sine  of  any  other 
small  arc. 
For  example,  find  the  sine  of  14'  21J";  that  is,  861"5. 

The  logarithmic  sine  of  1"  is  4.685576 

Add  logarithm  of  861.5,  2.935254 

Logarithmic  sine  of  14'  21£",  7.620830 

Two  lines  drawn,  the  one  from  the  surface  and  the 
other  from  the  center  of  the  earth,  to  the  center  of  the 
sun,  make  with  each  other  an  angle  of  8.61".  What  is 
the  logarithmic  sine  of  this  angle  ? 

The  log.  of  the  sine  1"  is  4.685575 

Log.  of  861,  0.935003 

Log.  sine  of  sun's  horizontal  parallax  =  5.620578 


GENERAL  APPLICATIONS  WITH  THE  USE  OF 
LOGARITHMS. 

I.  RIGHT-ANGLED  TRIGONOMETRY. 

One  figure  will  be  sufficient  to  represent  the  triangle 
in  all  of  the  following  examples ;  the  right  angle  being 
at£. 

PRACTICAL    PROBLEMS. 

1.  In  a  right-angled  triangle,  ABC, 
given  the  base  AB,  1214,  and  the  angle 
A,  51°  40'  30",  to  find  the  other  parts. 


SECTION    II.  289 

To  find  BC. 

Kadius,  10.000000 

:    tan.  .4,  51°  40'  30",  10.102119 

::    AB,12U,  3.084219 

:     BC,  1535.8,  3.186338 

Remark. — When  the  first  term  of  a  logarithmic  proportion  is  radius, 
the  required  logarithm  is  found  by  adding  the  second  and  third  loga- 
rithms, rejecting  10  in  the  index,  which  is  dividing  by  the  first  term. 

In  all  cases  we  add  the  second  and  third  logarithms  together ;  which, 
in  logarithms,  is  multiplying  these  terms  together ;  and  from  that  sum 
we  subtract  the  first  logarithm,  whatever  it  may  be,  which  is  dividing 
by  the  first  term. 

To  find  AC. 

Sin.  C,  or  cos.  A,  51°  40'  30",  9.792477 

:     AB,  1214,  3.084219 

::    Radius,  10.000000 

:    AC,  1957.7,  3.291742 

To  find  this  resulting  logarithm,  we  subtracted  the  first  logarithm 
from  the  second,  conceiving  its  index  to  be  13. 

Let  ABO  represent  any  plane  triangle,  right-angled 
at  B. 

2.  Given,  AO  73.26,  and  the  angle  A,  49°  12'  20"; 
required  the  other  parts. 

Ans.  The  angle  0, 40°  47'  40" ;  BO,  55.46 ;  and  AB,  47.87. 

8.  Given,  AB  469.34,  and  the  angle  A,  51°  26'  17",  to 
find  the  other  parts. 

Ans.  Theangletf,  38°  33' 43";  B  (7,588.7;  and^LC,  752.9. 

4.  Given,  AB  493,  and  the  angle  C,  20°  14' ;  required, 
the  remaining  parts. 

Ans.  The  angle  A,  69°  46';  BO,  1338 ;  and  AC,  1425.5. 

5.  Let  AB  =  331,  and  the  angle  A  =  49°  14' ;  what  are 
the  other  parts  ? 

Ans.  AC,  506.9;  BC,  383.9;  and  the  angle  O,  40°  46'. 

6.  If  AC=4:5,  and  the  angle  (7=37°  22',  what  are  the 
remaining  parts  ? 

Ans.  AB,  27.31 ;  BC,  35.76  ;  and  the  angle  A,  52°  38'. 
25  t 


290 


PLANE    TRIGONOMETRY. 


7.  Given,  AC=  4264.3,  and  the  angle  A  =  56°  29'  13", 
to  find  the  remaining  parts. 

Ans.AB,  2354.4;  BO,  3555.4;  and  the  angle  0,  33°  30'47". 

8.  If  AB  =  44.2,  and  the  angle  A  =  31°  12' 49",  what 
are  the  other  parts  ? 

Ans.  AC,  49.35 ;  BO,  25.57 ;  and  the  angle  0,  58°  47'  11". 

9.  If  ^LB  -  8372.1,  and  BO  =  694.73,  what  are  the 
other  parts  ? 

An8    (AC,  8400.9;   the  angle  0,   85°  15';    and  the 
'  \      angle  A,  4°  45'. 

10.  If  AB  be  63.4,  and  AC  be  85.72,  what  are  the 
other  parts? 

A       f  BO,  57.7 ;  the  angle  C,  47°  42';  and  the  angled, 
'  I     42°  18'. 

11.  Given,  AC  =  7269,  and  AB  =  3162,  to  find  the 
other  parts. 

A       (BO,  7546;  the  angle  C,  25°  47'  7";   and  the 
m'  \     angle  A,  64°  12'  53". 

12.  Given,  AC  =  4824,  and  BO  =  2412,  to  find  the 
other  parts. 

A       (  The  angle  A  =  30°  00',  the  angle  C  =  60°  00', 
m'  I      and  AB  -  4178. 

13.  The  distance  between  the  earth  and  sun  is  94,770,000 
miles,  and  at  that  distance  the  semi-diameter  of  the  sun 
subtends  an  angle  of  16'  6".  "What  is  the  diameter  of 
the  sun  in  miles  ?  Ans.  887,700  miles. 


In  this  example,  let  E  be  the  center  of  the  earth,  S  that  of  the 
sun,  and  EB  a  tangent  to  the  sun's  surface.  Then  the  A  EBS 
is  right-angled  at  B,  and  BJS  is  the  semi-diameter  of  the  sun.  The 
value  of  2BS  is  required. 


SECTION   II.  291 

14.  The  semi-diameter  of  the  earth  is  3956  miles,  and 
the  distance  of  the  sun  94.770000  miles.  What  angle 
will  the  semi-diameter  of  the  earth  subtend,  as  seen  from 
the  sun?  Ans.  8.60". 

This  angle  is  called,  in  astronomy,  the  sun's  horizontal  parallax. 
The  preceding  figure  applies  to  this  example,  by  supposing  E  to 
be  the  center  of  the  sun,  S  that  of  the  earth,' and  BS  equal  to 
3956  miles. 

15.  The  mean  distance  of  the  moon  from  the  earth  is 
60.3  times  3960  miles,  and  at.  this  distance  the  semi- 
diameter  of  the  moon  subtends  an  angle  of  15'  32". 
What  is  the  diameter  of  the  moon  in  miles  ? 

Ans.  2159  miles. 


H.   OBLIQUE-ANGLED  TRIGONOMETRY. 

PROBLEM   I. 

In  a  plane  triangle,  given  a  side  and  the  two  adjacent 
angles,  to  find  the  other  parts. 

'    In  the  triangle  ABC,  let  AB  =  c 

376,  the  angle  A  =  48°  3',  and  the 
angle  B  -  40°  14',  to  find  the  other 
parts. 

As  the  sum  of  the  three  angles  of  every  B 

triangle  is  always  180°,  the  third  angle,  C,  must  be  180°  —  88° 
17'  =  91°  43'. 


To  find  A  a 

Sin.  91°  43', 
:    AB,  376, 
: :  sin.  B  40°  14', 

9.999805 
2.575188 
9.810167 

12.385355 

:    .AC,  243, 

2.385550 

Observe,  that  the  sine  of  91°  43'  is  the  same  as  the  cosine  of 
1°  43'. 


292  PLANE    TRIGONOMETRY. 

To  find  BC. 

Sin.  91°  43',  9.999805 

:   A5,  376,  2.575188 

: :  sin.  A,  48°  3',  9.871414 

12.446602 


:   sin.  5  0,279.8,  2.446797 

PROBLEM   II. 

In  a  plane  triangle,  given  two  sides  and  an  angle  opposite 
one  of  them,  to  determine  the  other  parts. 

Let  AD  =  1751  feet,  one 
of  the  given  sides ;  the  angle 
D  =  31°  17'  19" ;  and  the  side 
opposite,  1257.5.  From  these 
data,  we  are  required  to  find 
the  other  side  and  the  other 
two  angles. 

In  this  case  we  do  not  know  whether  A  G  or  AE  represents 
1257.5,  because  AG  =  AE.  If  we  take  AG  for  the  other  given 
side,  then  D  G  is  the  other  required  side,  and  DA  G  is  the  vertical 
angle.  If  we  take  AE  for  the  other  given  side,  then  DE  is  the 
required  side,  and  DAE  is  the  vertical  angle.  In  such  cases  we 
determine  both  triangles. 

To  find  the  angle  U  =   G. 
(Prop.  4.)  AG  =  AE=  1257.5,    log.     3.099508 

:    D,  31°  17'  19",  sin.     9.715460 

: :  AD,  1751,  log.     3.243286 

12.958746 


<    E  =  G,  46°  18',  sin.     9.859238 

From  180°  take  46°  18',  and  the  remainder  is  the  angle  DGA 
=  133°  42'. 

The  angle     DAG  =  AGE —  D,  (Th.  11,  B.  I) y 
that  is,         DAG  =  46°  18'  —  31°  17'  19"  =  15°  0'41". 
The  angles  D  and  E,  taken  from  180°,  give 
DAE  =  102°  24'  41". 


SECTION    II.  293 

To  find  DO. 

Sin.  D,  31°  17'  19",  log.  9.715460 

:    ^1(7,  1257.5,  log.  3.099508 

: :  sin. DAG  15°  0'  41",  log.  0.413317 


12.512825 


:   DC,  626.86,  2.797165 

To  find  DE. 

Sin.  D,  31°  17' 17",  9.715460 

:    AE,  1257.5,  3.099508 

: :  sin.  DAE,  102°  24'  41",  9.989730 

13.089238 


:  DE,  2364.7,  3.373778 

Remark. — To  make  the  triangle  possible,  A  C  must  not  be  less  than 
AB,  the  sine  of  the  angle  D,  when  DA  is  made  radius. 

PROBLEM   III. 

In  any  plane  triangle,  given  two  sides  and  the  included 
angle,  to  find  the  other  parts. 

Let  AD  =  1751,  (see  last  figure),  DE  =  2364.5,  and 
the  included  angle  D  =  31°  17'  19".  "We  are  required 
to  find  AE,  the  angle  DAE,  and  the  angle  E. 

Observe  that  the  angle  E  must  be  less  than  the  angle  DAE,  be- 
cause it  is  opposite  a  less  side. 

From  180° 

Take  D,  31°  17'  19", 


Sum  of  the  other  two  angles,  =  148°  42'  41",  (Th.  11,  B.  I), 
*  sum  =     74°  21'  20". 

By  Proposition  7, 
DE+DA  :  DE—  DA  =  tan.  74°  21'  20"  :  tan.  \{DAE—  E). 
That  is, 

4115.5  :  613.5  =  tan.  74°  21' 20"  :  *a*k(DAE—  JP> 
25* 


294 


PLANE    TRIGONOMETRY. 


Tan.  74°  21'  20", 
613.5, 


4115.5  log.  (subtracted), 


10.552778 

2.787815 

13.340593 
3.614423 


tMi.i(DAEE-)  tan.28°  1'  36",  9.726170 

But  the  half  sum  plus  the  half  difference  of  any  two  quantities 

is  equal  to  the  greater  of  the  two;  and  the  half  sum  minus  the 

half  difference  is  equal  the  less. 

Therefore,  to  74°  21'  20", 


Add 


28°    V  36", 


DAE  = 
E        = 

102°  22'  56", 
46°  19'  45", 

To  find  AE. 

Sin.  E,  46°  19'  45", 
:    DA,  1751, 
: :  sin.  D,  31°  17'  19", 

:   XE,  1257.2, 

9.859323 
3.243286 
9.715460 

12.958746 

3.099423 

PROBLEM    IV. 

Given,  the  three  sides  of  a  plane  triangle,  to  find  the  angles. 

Let  AQ  m  1751,  OB  =  1257.5,  AB  =  2364.5,  to  find 
the  angles  A,  B,  and  O. 

If  we  take  the  formula  for  cosines,  we 
will  compute  the  greatest  angle,  which  is 
C.     To  correspond  with  the  formula, 


cos. 


0 


ab        '  A  B 

we  must  take     a  =  1257.5,  b  =  1751,  and  c  =  2364.5. 
The  half  sum  of  these  is, 

s  ==  2686.5;  and  s  —  c  =  322. 


SECTION    J 

[I. 

R* 

20.000000 

s  =  2686.5 

3.429187 

s  — c  =  322 

2.507856 

Numerator,  log. 

25.937043 

a  1257.5     3.099508 

b  1751.       3.243286 

295 


Denominator,  log.    6.342794       6.342794 

2)19.594249     ' 

\C=*    51°  11'  10"    cos.  9.797124 
C=102    22    20 

The  remaining  angles  may  now  be  found  by  Problem  4. 


PRACTICAL    PROBLEMS. 

Let  ABC  represent  any  oblique-angled  triangle. 

1.  Given,  AB  697,  the  angle  A  81°  30'  10",  and  the 
angle  B  40°  30'  44",  to  find  the  other  parts. 

Am.  AC,  534;  BC,  813;  and  [__C,  57°  59'  6". 

2.  If  A C  =  720.8,  [_A=7Q°  5'  22",  [__B  m  59°  35' 
36",  required  the  other  parts. 

Am.  AB,  643.2;  BC,  785.8;  and  [_C,  50°  19'  2". 

3.  Given,  BC  980.1,  the  angle  A  7°  6'  26",  and  the 
angle  B  106°  2'  23",  to  find  the  other  parts. 

Am.  AB,  7284;  AC,  7613.3;  and  [_C,  66°  51'  11". 

4.  Given,  AB  896.2,  B C  328.4,  and  the  angle   (7113° 
45'  20",  to  find  the  other  parts. 

A(  AC,  712;  LA  19°  35' 48"; 
'  I     and  [_B,  46°  38'  52". 

5.  Given,  AC r  =  4627,  BC=  5169,  and  the  angle  A  = 
70°  25'  12",  to  find  the  other  parts. 

A       (AB,  4328;  L^,  57°  29' 56"; 
'  \      and  [_C,  52°  4'  52". 


296  PLANE    TRIGONOMETRY. 

6.  Given,  AB  793.8,  BO  481.6,  and  AO  500.0,  to  find 
the  angles. 

AnjLA,  35°  15'  32";   IB,  36°  49'  18";   and   [__0, 
•\     107°  55'  10". 

7.  Given,  .45  100.3,  5(7  100.3,  and  AC  100.3,  to  find 
the  angles. 

A      i  The   angle  A,  60°;   the  angle  5,  60°;   and  the 
m'  \     angle  (7,  60°. 

8.  Given,  AB  92.6,  5(7  46.3,  and  AC  71.2,  to  find  the 
angles. 

,       f  LA  29°  IT'  22";    L*>  48°  47'  31";   and   [_0, 
^'1      101°  55'  8". 

9.  Given,  AB  4693,  BO  5124,  and  AO  5621,  to  find 
the  angles. 

A      j  [_A,  57°  30'  28";    \_B,  67°  42'  36";   and   [__0, 
^nS'\      54°  46'  56". 

10.  Given,  AB  728.1,  BO  614.7,  and  JL(7  583.8,  to  find 
the  angles. 

A      ]  \jkm  54°  32'  52",  \_B  =  50°  40'  58",  and  \__0 
#H      =74°  46'  10". 

11.  Given,  AB  96.74,  BO  83.29,  and  AO  111.42,  to 
find  the  angles. 

A      j  L^  =  46°  30'  45",    [_B  =  76°  3'  45",  and   [_0 
^'\      =57°  25'  30". 

12.  Given,  AB  363.4,  BO  148.4,  and  the  angle  B  102° 
18'  27",  to  find  the  other  parts. 

A       ( [_A  =  20°  9'  17",  the  side  A  0  =  420.8,  and  [__0 

13.  Given,  .45  632,  5(7  494,  and  the  angle  A  20°  16', 
to  find  the  other  parts,  the  angle  0  being  acute. 

([_ (7=  26°  18'  19",  [__B  =  133°  25'  41",  and 
ns'\     ^4(7=1035.86. 

14.  Given,  AB  53.9,  .4(7  46.21,  and  the  angle  B  58° 
16',  to  find  the  other  parts. 

Arts.  \__A  =  38°  58',  [__0=  82°  46',  and     5(7=  34.16. 


SECTION    II.  29T 

15.  Given,  AB  2163,  BC  1672,  and  the  angle  0  112° 
18'  22",  to  find  the  other  parts. 

Ana.  AC,  877.2;  [_B,  22°  2'  16";  and  [_A,  45°  39'  22". 

16.  Given,  AB  496,  BC  496,  and  the  angle  B  S8°  16', 
to  find  the  other  parts. 

Ans.  AC,  325.1;  [_A,  70°  52';  and  [__<?,  70°  52'. 

17.  Given,  AB  428,  the  angle  C  49°  16',  and  (AC+ 
BC)  918,  to  find  the  other  parts,  the  angle  B  being 
obtuse. 

A      (  The  angle  A  =  38°  44'  48",  the  angle  B  =  91° 
f*  \      59'  iSP^jitiL  564.49,  and  BC=  353.5. 

18.  Given,  AC  126,  the  angle  B  29°  46',  and  (AB— 
BC)  43,  to  find  the  other  parts. 

A      /The  angled  =  55°  51'  32",  the  angle  (7=94° 
**  I      22'  28",  AB  =  253.05,  and  BC=  210.054. 

19.  Given,  AB  1269,  .4(7  1837,  and  the  angle  A  53° 
16'  20",  to  find  the  other  parts. 

(  [_B  =  83°  23'  47",  L^=4^°  I9'  53",  and  BC 
AnS'\      =1482.16. 


298  PLANE    TRIGONOMETRY. 


SECTION   III 


APPLICATION  OF  TRIGONOMETRY  TO  MEASURING 
HEIGHTS  AND  DISTANCES. 

In  this  useful  application  of  Trigonometry,  a  base  line 
is  always  supposed  to  be  measured,  or  given  in  length ; 
and  by  means  of  a  quadrant,  sextant,  circle,  theodolite, 
or  some  other  instrument  for  measuring  angles,  such 
angles  are  measured  as,  connected  with  the  base  line  and 
the  objects  whose  heights  or  distances  it  is  proposed  to 
determine,  enable  us  to  compute,  from  the  principles  of 
Trigonometry,  what  those  heights  or  distances  are. 

Sometimes,  particularly  in  marine  surveying,  horizontal 
angles  are  determined  by  the  compass ;  but  the  varying 
effect  of  surrounding  bodies  on  the  needle,  even  in  situa- 
tions little  removed  from  each  other,  and  the  general 
construction  of  the  instrument  itself,  render  it  unfit  to  be 
employed  in  the  determination  of  angles  where  anything 
like  precision  is  required. 

The  following  problems  present  sufficient  variety,  to 
guide  the  student  in  determining  what  will  be  the  most 
eligible  mode  of  proceeding,  in  any  case  that  is  likely  to 
occur  in  practice. 

PROBLEM   I. 

Being  desirous  of  finding  the  distance  between  two 
distant  objects,  O  and  D,  I  measured  a  base,  AB,  of  384 
yards,  on  the  same  horizontal  plane  with  the  objects  0 


SECTION    III.  299 

and  D.  At  A,  I  found  the  angles  DAB  =  48°  12',  and 
CAB  =89°  18';  at  B,  the  angles  ABO  46°  14',  and 
ABB  87°  4'.  It  is  required,  from  these  data,  to  com- 
pute the  distance  between  C  and  B. 

From  the  angle  GAB,  take  the  angle  DAB ;  the 
remainder,  41°  6',  is  the  angle  CAD.  To  the  angle 
DBA,  add  the  angle  DAB,  and  44°  44',  the  supple- 
ment of  the  sum,  is  the  angle  ADB.  In  the  same 
way  the  angle  ACB,  which  is  the  supplement  of 
the  sum  of  CAB  and  CBA,  is  found  to  be  44°  28'.    A~~ 

Hence,  in  the  triangles  ABC  and  ABD,  we  have 

Sin.  ACB,  44°  28',  9.845405 

:   AB,  384  yards,  2.584331 

::  sin.  ABC,  46°  14',  9.858635 


12.442966 

:   AC,  395.9  yards, 

2.597561 

Sin.  ADB,  44°  44', 
:   AB,  384  yards, 
::  sin.  ABD,  87°  4', 

9.847454 
2.584331 
9.999431 

12.583762 

:   AD,  544.9  yards, 

2.736308 

Then,  in  the  triangle  CAD,  we  have  given  the  sides  0 A  and  AD, 
and  the  included  angle  CAD,  to  find  CD;  to  compute  which  we 
proceed  thus : 

The  supplement  of  the  angle  CAD,  is  the  sum  of  the  angles 
A  CD  and  ADC; 

_  ACD  +  ADC      aQO  _,        .    '  r 

Hence, =  o9°  2r;  and,  by  proportion  we  have, 

AD  +  AC  (=      940.8)        2.937497 

:  AD— AC  Cm         149)        2.173186 

::  \zn.ACD  +  AD°     (-  69°  27')      10.426108 


12.599294 


300 


PLANE   TRIGONOMETRY. 


tan. 


ACD—ADC 


(=  22  54)        9.551797 


the  angle  A  CD,     sum,  92  21 

the  angle  ADC,     diff.,  46  33 

Sin.  ADC,  46°  33', 
:    AC,  395.9  yards, 
: :  sin.  CAD,  41°  6', 


CD,  358.5  yards, 


9.860922 
2.597585 
9.817813 

12.415398 

2.554476 


PROBLEM   II. 

To  determine  the  altitude  of  a  lighthouse,  I  observed 
the  elevation  of  its  top  above  the  level  sand  on  the  sea- 
shore, to  be  15°  32'  18" ;  and  measuring  directly  from 
it,  638  yards  along  the  sand,  I  then  found  its  elevation 
to  be  9°  56'  26".    Required  the  height  of  the  lighthouse. 

Let  CD  represent  the  height  of  the  light- 
house above  the  level  of  the  sand,  and  let  B 
be  the  first  station,  and  A  the  second ;  then 
the  angle  CBD  is  15°  32'  18,  and  the  angle 
CAB  is  9°  56'  26";  therefore,  the  angle 
A  CB,  which  is  the  difference  of  the  angles 
CBD  and  CAB,  is  5°  35'  52". 

Hence,  Sin.  A  CB,  5°  35'  52", 

:    AB,  638, 
: :  sin.  angle  A,  9°  56'  26", 


8.989201 
2.804821 
9.237107 


BC,  1129.06  yards, 

Radius, 

BC,  1129.06, 

sin.  CBD,  15°  32'  18", 


DC,  302.46  yards, 


12.041928 
3.052727 

10.000000 
3.052727 
9.427945 

12.480672 

2.480672 


SECTION   III. 


301 


PROBLEM    III. 

Coming  from  sea,  at  the  point  B  I  observed  two 
headlands,  A  and  B,  and  inland,  at  0,  a  steeple,  which 
appeared  between  the  headlands.  I  found,  from  a  map, 
that  the  headlands  were  5.35  from  each  other;  that  the 
distance  from  A  to  the  steeple  was  2.8  miles,  and  from 
B  to  the  steeple  3.47  miles ;  and  I  found,  with  a  sextant, 
that  the  angle  ABO  was  12°  15,  and  the  angle  BBC,  15° 
30'.  Kequired  my  distance  from  each  of  the  headlands, 
and  from  the  steeple. 


CONSTRUCTION. 

The  angle  between  the  two  headlands  is 
the  sum  of  15°  30'  and  12°  15',  or  27°  45'. 
Take  double  this  sum,  55°  30'.  Conceive  AB 
to  be  the  chord  of  a  circle,  and  the  arc  on 
one  side  of  it  to  be  55°  30' ;  and,  of  course, 
the  other  will  be  304°  30'.  The  point  B 
will  be  somewhere  in  the  circumference  of 
this  circle.     Consider  that  point  as  determined,  and  draw  CB. 

In  the  triangle  ABO,  we  have  all  the  sides,  and,  of  course,  we 
can  find  all  the  angles;  and  if  the  angle  A  OB  is  less  than  180° — 
27°  45'  =  152°  15',  then  the  circle  cuts  the  line  CB  in  a  point 
E,  and  C  is  without  the  circle. 

Draw  AE,  BE,  AB,  and  BB.  AEBB  is  a  quadrilateral  in  a 
circle,  and  \__  AEB  +  [_  ABB  =  180°. 

The  [__  ABE  =  the  | ABE,  because  both  are  measured  by  one 

half  the  arc  AE.     Also,  | EBB  =  [__  EAB,  for  a  similar  reason. 

Now,  in  the  triangle  AEB,  its  side  AB,  and  all  its  angles,  are 
known ;  and  from  thence  AE  can  be  computed.  Then,  having  the 
two  sides,  AC  and  AE,  of  the  triangle  AEC,  and  the  included 
angle  CAE,  we  can  find  the  angle  AEC,  and,  of  course,  its  supple- 
ment, AEB.  Then,  in  the  triangle  AEB,  we  have  the  side  AE, 
and  the  two  angles  AEB  and  ABE,  from  which  we  can  find  AB. 

The  computation,  at  length,  is  as  follows : 
26 


302 


PLANE    TRIGONOMETRY. 


To  find  AJE. 

Angle  EAB  =     15°  30'  Sin.  AEB,  152°  15',  9.668027 

Angle  EBA  =     12°  15'  :    AB}  5.35,                    .728354 

27°  45'  : :  sin.  ABE  12°  15'      9.326700 

180° 


Angle  AEB  =  152°  15'       :   AE9  2.438, 


To  find  the  angle  BAG. 


10.055054 


.387027 


BG,    3.47 

AB,  5.35 

AC,  2.80 

log.     .728354 
log.     .447158 

2  )  11.62 

1.175512 

5.81 
BG,    2.34 

log.     .764176 
log.     .369216 
20 

21.133392 

17°  41'  58" 

2  )  19.957880 
cos.  9.978940 

2 

Anglo  BAG=  35°  23' 56" 
Angle  EAB  =  15°  30' 

Angle  EAG=  19°  53' 56" 
180° 

2  )  160°    6'    4" 
80°    3'    2" 

AEC+  AGE 

SECTION   III.  303 

To  find  the  angles  AEC  and  ACE. 


AC  +  AE 

:   AC  —  AE 

AEC  +  ACE 
::tan.             | 

5.238 
.362 

80°    3'    2" 
21°  30'  12" 

.719165 
—  1.558709 

10.755928 

10.314637 

AEC—  ACE 
:    tan.             g 

9.595472 

angle  .4.27(7,              101°  33/ 14",  sum 

angle  ACE  or  ACD,  58°  32' 50",  diff. 
angle  CD  A,  12° 15' 


70°  47' 50",  supplement  109°  12' 10",  angle  CAD 
35o  2y  56„  angle  CAB 

73°  48/ 14",  angle  BAD 


To  find  AD. 

Sin.  ADC,  12°15';  9.326700 

:   AC,2.S,  .447158 

::  sin.  4  CD  58°  32'  50",  9.930985 

10.378143 


:    AD  11.26  miles.  1.051443 


PROBLEM   IV. 


The  elevation  of  a  spire  at  one  station  was  23°  50'  17", 
and  the  horizontal  angle  at  this  station,  between  the 
spire  and  another  station,  was  93°  4'  20".  The  horizon- 
tal angle  at  the  latter  station,  between  the  spire  and  the 
first  station,  was  54°  28'  36",  and  the  distance  between 
the  two  stations  was  416  feet.  Required  the  height  of 
the  spire. 


304 


PLANE    TRIGONOMETRY. 


Let  CD  be  the  spire,  A  the  first  station,  and 
B  the  second ;  then  the  vertical  angle  CAD  is 
23°  50'  17";  and  as  the  horizontal  angles,  CAB 
and  CBA,  are  93°  4'  20"  and  54°  28'  36",  re-  B 
spectively,  the  angle  ACB,  the  supplement  of 
their  sum,  is  32°  27'  4". 


To  find  AC. 
Sin.BCA,  32°  27' 3", 
:    side  AB,  416, 
::  tan.  ABC,  54°  28' 36", 


:    side  AC,  631, 


9.729634 
2.619093 
9.910560 

12.529653 

2.800019 


To  find  DC. 

Radius,  10.000000 

side^Ltf,  631,  2.800019 

tan.  DA  C,  23°  50'  17",  9.645270 

DC,  278.8,  2.445289 


By  the  application  of  Pro- 
blem 4,  we  can  compute  the 
distance  between  two  horizon- 
tal planes,  if  the  same  object 
is  visible  from  both.  a^ 

For  example,  let  M  be  a 
prominent  tree  or  rock  near 

the  top  of  a  mountain,  and  by  observations  taken  at  A, 
we  can  determine  the  perpendicular  Mn.  By  like  obser- 
vations taken  at  B,  we  can  determine  the  perpendicular 
Mm.  The  difference  between  these  two  perpendiculars  is 
nm,  or  the  difference  in  the  elevation  between  the  two 
points  A  and  B.  If  the  distances  between  A  and  n,  or  B 
and  m,  are  considerable,  or  more  than  two  or  three  miles, 
corrections  must  be  made  for  the  convexity  of  the  earth ; 
but  for  less  distances  such  corrections  are  not  necessary. 


SECTION   III.  305 

PRACTICAL    PROBLEMS. 

1.  Eequired  the  height  of  a  wall  whose  angle  of  eleva- 
tion, at  the  distance  of  463  feet,  is  observed  to  be  16° 
21'.  Ans.  135.8  feet. 

2.  The  angle  of  elevation  of  a  hill  is,  near  its  bottom, 
31°  18',  and  214  yards  further  off,  26°  18'.  Eequired  the 
perpendicular  height  of  the  hill,  and  the  distance  of  the 
perpendicular  from  the  first  station. 

( The  height  of  the  hill  is  565.2  yards,  and  the 
Ans.  <      distance  of  the  perpendicular  from  the  first 
I     station  is  929.6  yards. 

3.  The  wall  of  a  tower  which  is  149.5  feet  in  height, 
makes,  with  a  line  drawn  from  the  top  of  it  to  a  distant 
object  on  the  horizontal  plane,  an  angle  of  57°  21'. 
"What  is  the  distance  of  the  object  from  the  bottom  of 
the  tower?  Ans.  233.3  feet. 

4.  From  the  top  of  a  tower,  which  is  138  feet  in  height, 
I  took  the  angle  of  depression  of  two  objects  standing 
in  a  direct  line  from  the  bottom  of  the  tower,  and  upon 
the  same  horizontal  plane  with  it.  The  depression  of  the 
nearer  object  was  found  to  be  48°  10',  and  that  of  the 
further,  18°  52'.  What  was  the  distance  of  each  from 
the  bottom  of  the  tower  ? 

*  .        (  Distance  of  the  nearer,  123.5  feet ;  and  of  the 
\      further,  403.8  feet. 

5.  Being  on  the  side  of  a  river,  and  wishing  to  know 
the  distance  of  a  house  on  the  opposite  side,  I  measured 
312  yards  in  a  right  line  by  the  side  of  the  river,  and  then 
found  that  the  two  angles,  one  at  each  end  of  this  line, 
subtended  by  the  other  end  and  the  house,  were  31°  15' 
and  86°  27'.  What  was  the  distance  between  each  end 
of  the  line  and  the  house  ?      Ans.  351.7,  and  182.8  yards. 

6.  Having  measured  a  base  of  260  yards  in  a  straight 
line,  on  one  bank  of  a  river,  I  found  that  the  two 
angles,  one  at  each  end  of  the  line,  subtended  by  the 

26*  u 


306  PLANE    TRIGONOMETRY. 

other  end  and  a  tree  on  the  opposite  bank,  were  40°  and 
80°.     What  was  the  width  of  the  river  ? 

Ans.  190.1  yards. 

7.  From  an  eminence  of  268  feet  in  perpendicular 
height,  the  angle  of  depression  of  the  top  of  a  steeple 
which  stood  on  the  same  horizontal  plane,  was  found  to 
be  40°  3',  and  of  the  bottom,  56°  18'.  "What  was  the 
height  of  the  steeple  ?  Ans.  117.8  feet. 

8.  Wanting  to  know  the  distance  between  two  objects 
which  were  separated  by  a  morass,  I  measured  the  dis- 
tance from  each  to  a  point  from  whence  both  could  be 
seen ;  the  distances  were  1840  and  1428  yards,  and  the 
angle  which,  at  that  point,  the  objects  subtended,  was  36° 
18'  24".    Required  their  distance.     Ans.  1090.85  yards. 

9.  From  the  top  of  a  mountain,  three  miles  in  height, 
the  visible  horizon  appeared  depressed  2°  13'  27".  Re- 
quired the  diameter  of  the  earth,  and  the  distance  of  the 
boundary  of  the  visible  horizon. 

a        f  Diameter  of  the  earth,  7958  miles ;  distance  of 
\      the  horizon,  154.54  miles. 

10.  From  a  ship  a  headland  was  seen,  bearing  north 
39°  23'  east.  After  sailing  20  miles  north,  47°  49'  west, 
.the  same  headland  was  observed  to  bear  north,  87°  11' 
east.  Required  the  distance  of  the  headland  from  the 
ship  at  each  station. 

A        /At  first  station,  19.09  miles ;   at  the  second, 
JLn8'  \     26.96  miles. 

11.  The  top  of  a  tower,  100  feet  above  the  level  of  the 
sea,  was  seen  as  on  the  surface  of  the  sea,  from  the  mast- 
head of  a  ship,  90  feet  above  the  water.  The  diameter 
of  the  earth  being  7960  miles,  what  was  the  distance 
between  the  observer  and  the  object? 

Ans.  23.9  plus  >fc  for  refraction  =  25.7  miles. 

12.  From  the  top  of  a  tower,  by  the  seaside,  143  feet 
high,  it  was  observed  that  the  angle  of  depression  of  a 


SECTION   III.  307 

ship's  bottom,  then  at  anchor,  measured  35°  ;  what,  then, 
was  the  ship's  distance  from  the  foot  of  the  tower  ? 

Arts.  204.22  feet. 

13.  Wanting  to  know  the  "breadth  of  a  river,  I  meas- 
ured a  base  of  500  yards  in  a  straight  line  on  one  bank; 
and  at  each  end  of  this  line  I  found  the  angles  subtended 
by  the  other  end  and  a  tree  on  the  opposite  bank  of  the 
river,  to  be  53°  and  79°  12'.  What,  then,  was  the  per- 
pendicular breadth  of  the  river?        Ans.  529.48  yards. 

14.  What  is  the  perpendicular  height  of  a  hill,  its 
angle  of  elevation,  taken  at  the  bottom  of  it,  being  46°, 
and  200  yards  further  off,  on  a  level  with  the  bottom, 
31°  ?  Am.  286.28  yards. 

15.  Wanting  to  know  the  height  of  an  inaccessible 
tower,  at  the  least  accessible  distance  from  it,  on  the 
same  horizontal  plane,  I  found  its  angle  of  elevation  to 
be  58°  ;  then  going  300  feet  directly  from  it,  I  found  the 
angle  there  to  be  only  32° ;  required  the  height  of  the 
tower,  and  my  distance  from  it  at  the  first  station. 

A       /Height,     307.53  feet. 
\  Distance,  192.15     " 

16.  Two  ships  of  war,  intending  to  cannonade  a  fort, 
are,  by  the  shallowness  of  the  water,  kept  so  far  from  it, 
that  they  suspect  their  guns  cannot  reach  it  with  effect. 
In  order,  therefore,  to  measure  the  distance,  they  separate 
from  each  other  a  quarter  of  a  mile,  or  440  yards,  and  then 
each  ship  observes  and  measures  the  angle  which  the 
other  ship  and  fort  subtends ;  these  angles  are  83°  45', 
and  85°  15'.  What,  then,  is  the  distance  between  each 
ship  and  the  fort  ?  ,        /  2292.26  yards. 

Jin8'  \  2298.05     " 

17.  A  point  of  land  was  observed  by  a  ship,  at  sea,  to 
bear  east-by-south  ;*  and  after  sailing  north-east  12  miles, 

*  That  is,  one  point  south  of  east.  A  point  of  the  compass  is 
11°  15'. 


308  PLANE    TRIGONOMETRY. 

it  was  found  to  bear  south-east-by-east.  It  is  required  to 
determine  the  place  of  that  headland,  and  the  ship's  dis- 
tance from  it  at  the  last  observation. 

Ans.  Distance,  26.0728  miles. 

18.  "Wishing  to  know  my  distance  from  an  inaccessible 
object,  0,  on  the  opposite  side  of  a  river,  and  having 
a  chain  or  chord  for  measuring  distances,  but  no  instru- 
ment for  taking  angles ;  from  each  of  two  stations,  A 
and  B,  which  were  taken  at  500  yards  asunder,  I  meas- 
ured in  a  direct  line  from  the  object,  0,  100  yards,  viz., 
A O  and  BD,  each  equal  to  100  yards;  and  I  found  that 
the  diagonal  AD  measured  550  yards,  and  the  diagonal 
BO  560.     What,  then,  was  the  distance  of  the  object  0 

from  each  station  A  and  B1     A         (  AO,  536.25  yards. 

Ji.ns.  1^500.09     " 

19.  A  navigator  found,  by  observation,  that  the  summit 
of  a  certain  mountain,  which  he  supposed  to  be  45  min- 
utes of  a  degree  distant,  had  an  altitude  above  the  sea 
horizon  of  31'  20".  Now,  on  the  supposition  that  the 
earth's  radius  is  3956  miles,  and  the  observer's  dip  was 
4'  15",  what  was  the  height  of  the  mountain  ? 

Ans.  3960  feet. 

Remark.  —  This  should  be  diminished  by  about  one  eleventh 
part  of  itself,  for  the  influence  of  horizontal  refraction. 

20.  From  two  ships,  A  and  B,  which  are  anchored  in 
a  bay,  two  objects,  O  and  i>,  on  the  shore,  can  be  seen. 
These  objects  are  known  to  be  500  yards  apart.  At  the 
ship  A,  the  angle  subtended  by  the  objects  was  measured, 
and  found  to  be  41°  25' ;  and  that  by  the  object  D  and 
the  other  ship  was  found  to  be  52°  12'.  At  the  other 
ship,  the  angle  subtended  by  the  objects  on  shore  was 
found  to  be  48°  10';  and  that  by  the  object  O,  and  the 
ship  A,  to  be  47°  40'.     Eequired  the  distance  between 


Ans. 


SECTION   III.  309 

the  ships,  and  the  distance  from  each  ship  to  the  objects 

on  shore. 

Distance  between  ships,     395.6  yards. 
From  ship  A  to  object  D,  743.5     " 
From  ship  A  to  object  0,  467.7     " 
.From  ship  B  to  object  D,  590.5     " 

To  solve  this  problem,  suppose  the  distance  between  the  ships  to 
be  100  yards,  and  determine  the  several  distances,  including  the 
distance  between  the  objects,  C  and  D,  under  this  supposition;  then 
multiply  the  values  thus  found  for  the  required  distances  by  the 
quotient  obtained  by  dividing  the  given  value  of  CD  by  the  com- 
puted value. 


PART    II. 

SPHERICAL   GEOMETRY 

AND 

TRIGONOMETRY. 


SECTION  I. 

SPHERICAL   GEOMETRY. 

DEFINITIONS. 

1.  Spherical  Geometry  has  for  its  object  the  investiga- 
tion of  the  properties,  and  of  the  relations  to  each  other, 
of  the  portions  of  the  surface  of  a  sphere  which  are 
bounded  by  the  arcs  of  its  great  circles. 

2.  A  Spherical  Polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs  of  great  circles,  called 
the  sides  of  the  polygon. 

3.  The  Angles  of  a  spherical  polygon  are  the  angles 
formed  by  the  bounding  arcs,  and  are  the  same  as  the 
angles  formed  by  the  planes  of  these  arcs. 

4.  A  Spherical  Triangle  is'a  spherical  polygon  having 
but  three  sides,  each  of  which  is  less  than  a  semi-circum- 
ference. 

5.  A  Lime  is  a  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  great  semi-circumferences  having  a 
common  diameter. 

6.  A  Spherical  Wedge,  or  TJngula,  is  a  portion  of  the 
surface  of  a  sphere  included  between  two  great  semi-cir- 
cles having  a  common  diameter. 

(310) 


SECTION   I.  3H 

7.  A  Spherical  Pyramid  is  a  portion  of  a  sphere  bounded 
by  the  faces  of  a  solid  angle  having  its  vertex  at  the 
center,  and  the  spherical  polygon  which  these  faces  inter- 
cept on  the  surface.  This  spherical  polygon  is  called  the 
base  of  the  pyramid. 

8.  The  Axis  of  a  great  circle  of  a  sphere  is  that  diameter 
of  the  sphere  which  is  perpendicular  to  the  plane  of  the 
circle.  This  diameter  is  also  the  axis  of  all  small  circles 
parallel  to  the  great  circle. 

9.  A  Pole  of  a  circle  of  a  sphere  is  a  point  on  the  sur- 
face of  the  sphere  equally  distant  from  every  point  in  the 
circumference  of  the  circle. 

10.  Supplemental,  or  Polar  Triangles,  are  two  triangles  on 
a  sphere,  so  related  that  the  vertices  of  the  angles  of 
either  triangle  are  the  poles  of  the  sides  of  the  other. 

PROPOSITION   I. 

Any  two  sides  of  a  spherical  triangle  are  together  greater 
than  the  third  side. 

Let  AB,  AC,  and  BO,  be  the  three 
sides  of  the  triangle,  and  D  the  center 
of  the  sphere. 

The  arcs  AB,  AC,  and  BO,  are  meas- 
ured by  the  angles  of  the  planes  that 
form  the  solid  angle  at  D.  But  any 
two  of  these  angles  are  together  greater 
than  the  third  angle,  (Th.  18,  B.  VI).  Therefore,  any  two 
sides  of  the  triangle  are,  together,  greater  than  the  third  side. 

Hence  the  proposition. 

PROPOSITION   II. 

The  sum  of  the  three  sides  of  any  spherical  triangle  is  less 
than  the  circumference  of  a  great  circle. 

Let  ABO  be  a  spherical  triangle ;.  the  two  sides,  AB 
and  A  0,  produced,  will  meet  at  the  point  which  is  diame- 
trically opposite  to  A,  and  the  arcs,  ABB  and  AOB  are 


312 


SPHERICAL    GEOMETRY. 


together  equal  to  a  great  circle.  But, 
by  the  last  proposition,  BO  is  less 
than  the  two  arcs,  BD  and D 0.  There- 
fore, AB  +  BC  +  AC,  is  less  than 
ABB  +  A  CD ;  that  is,  less  than  a 
great  circle. 

Hence  the  proposition. 


PROPOSITION   III. 

The  poles  of  a  great  circle  of  a  sphere  are  the  extremities 
of  its  axis,  and  these  points  are  also  the  poles  of  all  small 
circles  parallel  to  the  great  circle. 

Let  0  be  the  center  of 
the  sphere,  and  BD  the 
axis  of  the  great  circle, 
Cm  Am" ;  then  will  B  and 
D,  the  extremities  of  the 
axis,  be  the  poles  of  the 
circle,  and  also  the  poles 
of  any  parallel  small  cir- 
cle, as  FnJEJ. 

For,  since  BD  is  per- 
pendicular to  the  plane 
of  the  circle,  Cm  Am",  it 

is  perpendicular  to  the  lines  OA,  Om',  Om",  etc.,  passing 
through  its  foot  in  the  plane,  (Th.  3,  B.  VI);  hence,  all 
the  arcs,  Bm,  Bm',  etc.,  are  quadrants,  as  are  also  the 
arcs  Dm,  Dm1,  etc.  The  points  B  and  D  are,  therefore, 
each  equally  distant  from  all  the  points  in  the  circumfer- 
ence, Cm  Am" ;  hence,  (Def.  9),  they  are  its  poles. 

Again,  since  the  radius,  OB,  is  perpendicular  to  the 
plane  of  the  circle,  Cm  Am",  it  is  also  perpendicular  to 
the  plane  of  the  parallel  small  circle,  FnE,  and  passes 
through  its  center,  0f.  Now,  the  chords  of  the  arcs,  BF, 
Bn,  BF,  etc.,  being  oblique  lines,  meeting  the  plane  of 
the  small  circle  at  equal  distances  from  the  foot  of  the 


■p  y^- — 

0'~ --N.F 

-Jr 

»""■"""   i 

in 

m"             \ 

o        ^^A 

mi> 

SECTION    I.  313 

perpendicular,  BOf,  are  all  equal,  (Tli.  4,  B.  VI);  hence, 
the  arcs  themselves  are  equal,  and  B  is  one  pole  of  the 
circle,  FnF.  In  like  manner  we  prove  the  arcs,  BF,  Bn, 
BF,  etc.,  equal,  and  therefore  B  is  the  other  pole  of  the 
same  circle. 

Hence  the  proposition,  etc. 

Cor.  1.  A  point  on  the  surface  of  a  sphere  at  the  distance 
of  a  quadrant  from  tivo  points  in  the  arc  of  a  great  circle,  not 
at  the  extremities  of  a  diameter,  is  a  pole  of  that  arc. 

For,  if  the  arcs,  Bm,  Bm!,  are  each  quadrants,  the  angles, 
BOm  and  BOmf,  are  each  right  angles;  and  hence,  BO 
is  perpendicular  to  the  plane  of  the  lines,  Om  and  0m\ 
which  is  the  plane  of  the  arc,  m  m> f;  B  is  therefore  the 
pole  of  this  arc. 

Cor.  2.  The  angle  included  between  the  arc  of  a  great  circle 
and  the  arc  of  another  great  circle,  connecting  any  of  its  points 
with  the  pole,  is  a  right  angle. 

For,  since  the  radius,  BO,  is  perpendicular  to  the  plane 
of  the  circle,  Cm  Am",  every  plane  passed,  through  this 
radius  is  perpendicular  to  the  plane  of  the  circle ;  hence, 
the  plane  of  the  arc  Bm  is  perpendicular  to  that  of  the 
arc  Cm',  and  the  angle  of  the  arcs  is  that  of  their  planes. 

PROPOSITION    IV. 

The  angle  formed  by  two  arcs  of  great  circles  ivhich  inter- 
sect each  other,  is  equal  to  the  angle  included  between  the  tan- 
gents to  these  arcs  at  their  point  of  intersection,  and  is  meas- 
ured by  that  arc  of  a  great  circle  whose  pole  is  the  vertex  of 
the  angle,  which  is  limited  by  the  sides  of  the  angle  or  the 
sides  produced. 

Let  AM  and  AJSf  be  two  arcs  intersecting  at  the 
point  A,  and  let  AE  and  AF  be  the  tangents  to  these 
arcs  at  this  point.  Take  A  C  and  AB,  each  quadrants, 
and  draw  the  arc  CD,  of  which  A  is  the  pole,  and  00 
and  OB  are  the  radii. 
27 


314 


SPHERICAL   GEOMETRY. 


!Now,  since  the  planes  of  the  arcs  intersect  in  the  radius 
OA,  and  AE  is  a  tangent  to  one  arc,  and  AF  a  tangent 
to  the  other,  at  the  common  point  A, 
these  tangents  form  with  each  other  an 
angle  which  is  the  measure  of  the  angle 
of  the  planes  of  the  arcs ;  but  the  angle 
of  the  planes  of  the  arcs  is  taken  as  the 
angle  included  by  the  arcs,  (Def.  4). 

Again,  because  the  arcs,  AC  and  AD, 
are  each  quadrants,  the  angles,  A  00, 
A  OD,  are  right  angles ;  hence  the  radii, 
00  and  OD,  lie,  the  one  in  one  face, 
and  the  other  in  the  other  face,  of  the 
diedral  angle  formed  by  the  planes  of  the  arcs,  and  are 
perpendicular  to  the  common  intersection  of  these  faces 
at  the  same  point.  The  angle,  OOD,  is  therefore  the 
angle  of  the  planes,  and  consequently  the  angle  of  the 
arcs  ;  but  the  angle  OOD  is  measured  by  the  arc  OD, 

Hence  the  proposition. 

Oor.  1.  Since  the  angles  included  between  the  arcs  of 
great  circles  on  a  sphere,  are  measured  by  other  arcs  of 
great  circles  of  the  same  sphere,  we  may  compare  such 
angles  with  each  other,  and  construct  angles  equal  to 
other  angles,  by  processes  which  do  not  differ  in  principle 
from  those  by  which  plane  angles  are  compared  and  con- 
structed. 

Oor.  2.  Two  arcs  of  great  circles  will  form,  by  their  in- 
tersection, four  angles,  the  opposite  or  vertical  ones  of 
which  will  be  equal,  as  in  the  case  of  the  angles  formed 
by  the  intersection  of  straight  lines,  (Th.  4,  B.  I). 


PROPOSITION  V. 

The  surface  of  a  hemisphere  may  be  divided  into  three  right- 
angled  and  four  quadrantal  triangles,  and  one  of  these  right- 
angled  triangles  will  be  so  related  to  the  other  two,  that  two 
of  its  sides  and  one  of  its  angles  will  be  complemental  to  the 


SECTION    I.  315 

Bides  of  one  of  them,  and  two  of  its  sides  supplemental  to  two 
of  the  sides  of  the  other. 

Let  ABO  be  a  right-angled  spherical  triangle,  right 
angled  at  B. 

Produce  the  sides,  AB  and  A  0,  and 
they  will  meet  at  A',  the  opposite 
point  on  the  sphere.  Produce  BO, 
both  ways,  90°  from  the  point  B,  to 
P  and  P',  which  are,  therefore,  poles 
to  the  arc  AB,  (Prop.  3).  Through 
A,  P,  and  the  center  of  the  sphere, 
pass  a  plane,  cutting  the  sphere  into 
two  equal  parts,  forming  a  great  circle  on  the  sphere, 
which  great  circle  will  be  represented  by  the  circle 
PAP' A1  in  the  figure.  At  right  angles  to  this  plane, 
pass  another  plane,  cutting  the  sphere  into  two  equal 
parts  ;  this  great  circle  is  represented  in  the  figure  by  the 
straight  line,  POP'.  A  and  A'  are  the  poles  to  the  great 
circle,  POP'  \  and  P  and  P'  are  the  poles  to  the  great 
circle,  ABA'. 

Now,  OPB  is  a  spherical  triangle,  right-angled  at  D, 
and  its  sides  OP  and  OB  are  complemental  respectively 
to  the  sides  £67and  AO of  the  A  ABO,  and  its  side  PB 
is  .complemental  to  the  arc  BO,  which  measures  the 
[_B A  0  of  the  same  triangle.  Again,  the  A  A'B 0  is  right- 
angled  at  B,  and  its  sides  A'O,  A'B,  are  supplemental 
respectively  to  the  sides  AO,  AB,  of  the  a  ABO.  There- 
fore, the  three  right-angled  A's,  ABO,  OPB,  and  A'BO, 
have  the  required  relations.  In  the  A  AOP,  the  side  AP 
is  a  quadrant,  and  for  this  reason  the  A  is  called  a  quad- 
ran  tal  triangle.  So  also,  are  the  A's  A' OP,  AOP',  and 
P'OA',  quadrantal  triangles.     Hence  the  proposition. 

Scholium. — In  every  triangle  there  are  six  elements,  three  sides  and 
three  angles,  called  the  parts  of  the  triangle. 

Now,  if  all  the  parts  of  the  triangle  ABC  are  known,  the  parts  of 
each  of  the  A's>  PCD  and  A'BC,  are  as  completely  known.  And 
when  the  parts  of  the  ^  PCD  are  known,  the  parts  of  the  A's  A  CP 


316  SPHERICALTRIGONOMETRY. 

and  A'CP  are  also  known  ;  for,  the  side  PD  measures  each  of  the  | 'a 

PAC  and  PA'C,  and  the  angle  CPD,  added  to  the  right  angle  A'PD, 

gives  the  [_A/PC,  and  the  | CPA  is  supplemental  to  this.     Hence, 

the  solution  of  the  A  ABC  is  a  solution  of  the  two  right-angled  and 
four  quadrantal  A's,  which  together  with  it  make  up  tne  surface  of 
the  hemisphere. 

PROPOSITION  VI. 

If  there  be  three  ares  of  great  circles  whose  poles  are  the 
angular  points  of  a  spherical  triangle,  such  arcs,  if  produced, 
will  form  another  triangle,  whose  sides  will  be  supplemental 
to  the  angles  of  the  first  triangle,  and  the  sides  of  the  first 
triangle  will  be  supplemental  to  the  angles  of  the  second. 

Let  the  arcs  of  the  three  great  cir- 
cles be  GH,  PQ,  KL,  whose  poles  are 
respectively  A,  B,  and  Q.  Produce  the 
three  arcs  until  they  meet  in  D,  E,  and 
F.  We  are  now  to  prove  that  E  is  the 
pole  of  the  arc  AO;  D  the  pole  of  the 
arc  BO;  F  the  pole  to  the  arc  A B. 
Also,  that  the  side  EF,  is  supplemental 
to  the  angle  A;  ED  to  the  angle  0; 
and  DF  to  the  angle  B;  and  also,  that  the  side  A  0  is 
supplemental  to  the  angle  E,  etc. 

A  pole  is  90°  from  any  point  on  the  circumference  of 
its  great  circle ;  and,  therefore,  as  A  is  the  pole  of  the 
arc  GH,  the  point  A  is  90°  from  the  point  E.  As  0  is 
the  pole  of  the  arc  LK,  0  is  90°  from  any  point  in 
that  arc;  therefore,  0  is  90°  from  the  point  E;  and 
E  being  90°  from  both  A  and  O,  it  is  the  pole  of  the  arc 
AQ.  In  the  same  manner,  we  may  prove  that  D  is  the 
pole  of  BO,  and  F  the  pole  of  AB. 

Because  A  is  the  pole  of  the  arc  G H,  the  arc  GH 
measures  the  angle  A,  (Prop.  4) ;  for  a  similar  reason, 
PQ  measures  the  angle  B,  and  LK  measures  the  angle  O. 

Because  E  is  the  pole  of  the  arc    AO,     EH  =  90° 

Or,  EG+GH=  90° 

For  a  like  reason,  FH  -f  GH  =  90° 


SECTION   I.  317 

Adding  these  two  equations,  and  observing  that  GH 
=  A,  and  afterward  transposing  one  A,  we  have, 
EG  +  GH+FH=1SQ°  —  A. 

Or,  EF=  180°— A  j) 

In  like  manner,  JF!Z>  =  180°  —  B    \     («) 

And,  BE  =  180°  —  0  J 

But  the  arc  (180°—  A),  is  a  supplemental  arc  to  A,  by 
the  definition  of  arcs ;  therefore,  the  three  sides  of  the 
triangle  BEF,  are  supplements  of  the  angles  A,  B,  0,  of 
the  triangle  ABO. 

Again,  as  E  is  the  pole  of  the  arc  A  (7,  the  whole  angle 
E  is  measured  by  the  whole  arc  LE. 

But,  AC  +  CH  =  90° 

Also,  AC  +  AL  =  90° 

By  addition,  AC+AC+CE  +  AL  =  180° 
By  transposition,  1(T+  CE+AL  =  180°— A  O 
That  is,  LIT,  or  .#=  180°— A O  ^ 

In  the  same  manner,  .F  =  180°— ^LB    M6) 

And,  E=1S0°  — BO  J 

That  is,  the  sides  of  the  first  triangle  are  supplemental 
to  the  angles  of  the  second  triangle. 


PKOPOSITION   VII. 

The  sum  of  the  three  angles  of  any  spherical  triangle,  is 
greater  than  two  right  angles,  and  less  than  six  right  angles. 

Add  equations  (a),  of  the  last  proposition.  The  first 
member  of  the  equation  so  formed  will  be  the  sum  of 
the  three  sides  of  a  spherical  triangle,  which  sum  we 
may  designate  by  S.  The  second  member  will  be  6  right 
angles  (there  being  2  right  angles  in  each  180°)  less  the 
three  angles  A,  B,  and  O. 

That  is,  S  =  6  right  angles  —  (A+B+C) 

By  Prop.  2,  the  sum  S  is  less  than  4  right  angles; 


318  SPHERICAL    GEOMETRY. 

therefore,  to  it  add  s,  a  sufficient  quantity  to  make  4 
right  angles.     Then, 

4  right  angles  =  6  right  angles  —  (A-hB+0)-\-s 

Drop  or  cancel  4  right  angles  from  both  members,  and 
transpose  (A  +  B  +  0). 

Then,  A  +  B  +  0  =  2  right  angles  +  s. 

That  is,  the  three  angles  of  a  spherical  triangle  make 
a  greater  sum  than  two  right  angles  by  the  indefinite 
quantity  s,  which  quantity  is  called  the  spherical  excess, 
and  is  greater  or  less  according  to  the  size  of  the  triangle. 

Again,  the  sum  of  the  angles  is  less  than  6  right  angles. 
There  are  but  three  angles  in  any  triangle,  and  each  one  of 
them  must  be  less  than  180°,  or  2  right  angles.  For,  an 
angle  is  the  inclination  of  two  lines  or  two  planes ;  and 
when  two  planes  incline  by  180°,  the  planes  are  parallel, 
or  are  in  one  and  the  same  plane ;  therefore,  as  neither 
angle  can  be  equal  to  2  right  angles,  the  three  can  never 
be  equal  to  6  right  angles. 

.      PROPOSITION    VIII. 

On  the  same  sphere,  or  on  equal  spheres,  triangles  which 
are  mutually  equilateral  are  also  mutually  equiangular ;  and, 
conversely,  triangles  which  are  mutually  equiangular  are  also 
mutually  equilateral,  equal  sides  lying  opposite  equal  angles. 

First— "Let  ABO  and  DBF,  in 
which  AB  =  BE,  AO=  DF,  and 

BO  =  EF,  be  two  triangles  on 
the  sphere  whose  center  is  0; 
then  will  the  [_  A,  opposite  the 
side  BO,  in  the  first  triangle,  be 
equal  the  [_JD,  opposite  the  equal 
side  EF,  in  the  second;  also 
l_B=[__E,  and  \__0=[_F. 


SECTION   I.  319 

For,  drawing  the  radii  to  the  vertices  of  the  angles  of 
these  triangles,  we  may  conceive  0  to  be  the  common 
vertex  of  two  triedral  angles,  one  of  which  is  hounded 
by  the  plane  angles  A  OB,  BOO,  and  A  00,  and  the  other 
by  the  plane  angles  DOE,  EOF,  and  DOE.  But  the 
plane  angles  bounding  the  one  of  these  triedral  angles, 
are  equal  to  the  plane  angles  bounding  the  other,  each 
to  each,  since  they  are  measured  by  the  equal  sides  of  the 
two  triangles.  The  planes  of  the  equal  arcs  in  the  two 
triangles  are  therefore  equally  inclined  to  each  other, 
(Th.  20,  B.  VI) ;  but  the  angles  included  between  the 
planes  of  the  arcs  are  equal  to  the  angles  formed  by  the 
arcs,  (Def.  3). 

Hence  the  [_A,  opposite  the  side  J?0,  in  the  A  ABO, 
is  equal  to  the  [__  B,  opposite  the  equal  side  EF,  in  the 
other  triangle ;  and  for  a  similar  reason,  the  [__B=  \__E, 
and  the  [_0=[_F. 

Second. — If,  in  the  triangles  ABO  and  BEE,  being  on 
the  same  sphere  whose  center  is  0,  the  |__  A  =  [__  B,  the 
[_B  =  [_E,  and  the  [_0=  [_E;  then  will  the  side  AB, 
opposite  the  [__  0>  in  *ne  first?  be  equal  to  the  side  BE, 
opposite  the  equal  L_  E,  in  the  second ;  and  also  the  side 
AO  equal  to  the  side  BE,  and  the  side  BO  equal  to  the 
side  EF. 

For,  conceive  two  triangles,  denoted  by  A'B'O'  and 
B'E'F',  supplemental  to  ABO  and  BEE,  to  be  formed; 
then  wTill  these  supplemental  triangles  be  mutually  equi- 
lateral, for  their  sides  are  measured  by  180°  less  the 
opposite  and  equal  angles  of  the  triangles  ABO  and 
BEF,  (Prop.  6) ;  and  being  mutually  equilateral,  they 
are,  as  proved  above,  mutually  equiangular.  But  the 
triangles  ABO  and  BEE  are  supplemental  to  the  tri- 
angles A'B'O'  and  B'E'F' ';  and  their  sides  are  therefore 
measured  severally  by^l80°  less  the  opposite  and  equal 
angles  of  the  triangles  A'B'O'  and  B'E'E',  (Prop.  6). 


320  SPHERICAL    GEOMETRY. 

Hence  the  triangles  ABO  and  BEF,  which  are  mutually 
equiangular,  are  also  mutually  equilateral. 

Scholium. — With  the  three  arcs  of  great  circles,  AB,  AC,  and  BC, 
either  of  the  two  triangles,  ABC,  BEF,  may  be  formed ;  hut  it  is  evi- 
dent that  these  two  triangles  cannot  be  made  to  coincide,  though  they 
are  both  mutually  equilateral  and  mutually  equiangular.  Spherical 
triangles  on  the  same  sphere,  or  on  equal  spheres,  in  which  the  sides 
and  angles  of  the  one  are  equal  to  the  sides  and  angles  of  the  other, 
each  to  each,  but  are  not  themselves  capable  of  superposition,  are 
called  symmetrical  triangles. 


PROPOSITION    IX. 

On  the  same  sphere,  or  on  equal  spheres,  triangles  having 
two  sides  of  the  one*  equal  to  two  sides  of  the  other,  each  to 
each,  and  the  included  angles  equal,  have  their  remaining 
sides  and  angles  equal. 

Let  ABO  and  DEF  be  two 
triangles,  in  which  AB  —  BE, 
AO =  BF,  and  the  angle  A  — 
the  angle  B ;  then  will  the  side 
BO  be  equal  to  the  side  FE, 
the  [_B  =  theL^andLtf 

=  L^. 

For,  if  BE  lies  on  the  same 
side  of  BF  that  AB  does  of  AO,  the  two  triangles,  ABO 
and  BEF,  may  be  applied  the  one  to  the  other,  and  they 
may  be  proved  to  coincide,  as  in  the  case  of  plane  tri- 
angles. But,  if  BE  does  not  lie  on  the  same  side  of  BF 
that  AB  does  of  AO,  we  may  construct  the  triangle  which 
is  symmetrical  with  BEF;  and  this  symmetrical  triangle, 
when  applied  to  the  triangle  ABO,  will  exactly  coincide 
with  it.  But  the  triangle  BEF,  and  the  triangle  sym- 
metrical with  it,  are  not  only  mutually  equilateral,  but 
also  are  mutually  equiangular,  the  equal  angles  lying 
opposite  the  equal  sides,  (Prop.  8) ;  and  as  the  one  or  the 
other  will  coincide  with  the  triangle  ABO,  it  follows  that 


SECTION    I.  321 

the  triangles,  ABC  and  BJEF,  are  either  absolutely  or 
symmetrically  equal. 

Cor.  On  the  same  sphere,  or  on  equal  spheres,  triangles 
having  two  angles  of  the  one  equal  to  two  angles  of  the  other, 
each  to  each,  and  the  included  sides  equal,  have  their  remain- 
ing sides  and  angles  equal. 

For,  if  [__A  =  L Pi  L.B  =  [_U,  and  side  AB  =  side 
BE,  the  triangle  BEF,  or  the  triangle  symmetrical  with 
it,  will  exactly  coincide  with  A  ABO,  when  applied  to  it 
as  in  the  case  of  plane  triangles ;  hence,  the  sides  and 
angles  of  the  one  will  be  equal  to  the  sides  and  angles 
of  the  other,  each  to  each. 


PROPOSITION   X. 

In  an  isosceles  spherical  triangle,  the  angles  opposite  the 
equal  sides  are  equal. 

A 

Let  ABO  be  an  isosceles  spherical  tri- 
angle, in  which  AB  and  A  0  are  the  equal 
sides ;  then  will  [__  B  =  Q  0. 

For,  connect  the  vertex  A  with  B,  the  i 

middle  point  of  the  base,  by  the  arc  of  a  / 
great  circle,  thus  forming  the  two  mutu-  ^4—. 
ally  equilateral  triangles,  ABB  and  ABO. 
They  are  mutually  equilateral,  because  AB  is  common, 
BB  =  DC  by  construction,  and  AB=AObj  supposition; 
hence  they  are  mutually  equiangular,  the  equal  angles 
being  opposite  the  equal  sides,  (Prop.  8).  The  angles  B 
and  0,  being  opposite  the  common  side  AB,  are  there- 
fore equal. 

Cor.  The  arc  of  a  great  circle  which  joins  the  vertex 
of  an  isosceles  spherical  triangle  with  the  middle  point  of 
the  base,  is  perpendicular  to  the  base,  and  bisects  the  ver- 
tical angle  of  the  triangle ;  and,  conversely,  the  arc  of  a 

v 


322 


SPHERICAL    GEOMETRY. 


great  circle  which  bisects  the  vertical  angle  of  an  isosceles 
spherical  triangle,  is  perpendicular  to,  and  bisects  the 
base. 


PROPOSITION   XI. 

If  two  angles  of  a  spherical  triangle  are  equal,  the  opposite 
sides  are  also  equal,  and  the  triangle  is  isosceles. 

In  the  spherical  triangle,  ABC,  let  the  \__B  =  [__C;  then 
will  the  sides,  AB  and  AC,  opposite  these  equal  angles, 
be  equal. 

For,  let  P  be  the  pole  of  the  base,  BO, 
and  draw  the  arcs  of  great  circles,  PB, 
PC;  these  arcs  will  be  quadrants,  and  at 
right  angles  to  BC,  (Cor.  1,  Prop.  3). 
Also,  produce  CA  and  BA  to  meet  PB 
and  PC,  in  the  points  E  and  F.  Now, 
the  angles,  PBF  and  PCE,  are  equal, 
because  the  first  is  equal  to  90°  less  the 
[_ABC,  and  the  second  is  equal  to  90° 
less  the  equal  [_ACB;  hence,  the  A's, 
PBF  and  PCE,  are  equal  in  all  their  parts, 
since  they  have  the  [_P  common,  the  \_PBF  =  [_PCE, 
and  the  side  PB  equal  to  the  side  PC,  (Cor.,  Prop.  9). 
PE  is  therefore  equal  to  PF,  and  [_PEC=  [__PFB. 

Taking  the  equals  PF  and  PE,  from  the  equals  PC 
and  PB,  we  have  the  remainders,  FC  and  EB,  equal ; 
and,  from  180°,  taking  the  [_'s  PFB  and  PEC,  we  have 
the  remaining  L_'s,  AFC  and  AEB,  equal.  Hence,  the 
A's,  AFC  and  AEB,  have  two  angles  of  the  one  equal  to 
two  angles  of  the  other,  each  to  each,  and  the  included 
sides  equal;  the  remaining  sides  and  angles  are  therefore 
equal,  (Cor.,  Prop.  9).  Therefore,  A  C  is  equal  to  BA> 
and  the  A  ABC  is  isosceles. 

Cor.  An  equiangular  spherical  triangle  is  also  equilat- 
eral, and  the  converse. 


SECTION    I.  323 

Remark.  —  In  this  demonstration,  the  pole  of  the  base,  BC,  is  sup- 
posed to  fall  without  the  triangle,  ABC.  The  same  figure  may  be  used 
for  the  case  in  which  the  pole  falls  within  the  triangle ;  the  modifi- 
cation the  demonstration  then  requires  is  so  slight  and  obvious,  that 
it  would  be  superfluous  to  suggest  it. 


PROPOSITION    XII. 

The  greater  of  two  sides  of  a  spherical  triangle  is  opposite 
the  greater  angle  ;  and,  conversely,  the  greater  of  two  angles 
of  a  spherical  triangle  is  opposite  the  greater  side. 

Let  ABO  be  a  spherical  triangle,  in  which  the  angle  A 
is  greater  than  the  angle  B;  then  is  the  side  BO  greater 
than  the  side  A  0. 

Through  A  draw  the  arc  of  a 
great  circle,  AD,  making,  with  AB, 
the  angle  BAB  equal  to  the  angle 
ABB.  The  triangle,  BAB,  is  isos- 
celes, and  DA  =  BB,  (Prop.  11). 

In  the  A  AOD,  AO<  OD  +  AD, 
(Prop.  1) ;  or,  substituting  for  AD  its  equal  DB,  we  have, 

AO  <  OB  +  DB. 

Inverting  the  members  of  the  inequality,  and  writing 
OB  for  OB  +  DB,  it  becomes  OB  >  OA. 

Conversely ;  if  the  side  OB  be  greater  than  the  side  OA, 
then  is  the  [_A  >  the  [_B.  For,  if  the  [_A  is  not  greater 
than  the  [__B,  it  is  either  equal  to  it,  or  less  than  it.  The 
\_A  is  not  equal  to  the  [_B ;  for  if  it  were,  the  triangle 
would  be  isosceles,  and  OB  would  be  equal  to  OA,  which 
is  contrary  to  the  hypothesis.  The  [_A  is  not  less  than 
the  [___B;  for  if  it  were,  the  side  OB  would  be  less  than  the 
side  OA,  by  the  first  part  of  the  proposition,  which  is  also 
contrary  to  the  hypothesis ;  hence,  the  [_A  must  be  greater 
than  the  L^. 


324  SPHERICAL    GEOMETRY. 

PROPOSITION    XIII. 
Two  symmetrical  spherical  triangles  are  equal  in  area. 

Let  ABO  and  DEF  be  two  A's  on  the  same  sphere, 
having  the  sides  and  angles  of  the  one  equal  to  the  sides 
and  angles  of  the  other,  each  to 
each,  the  triangles  themselves 
not  admitting  of  superposition. 
It  is  to  be  proved  that  these 
A's  have  equal  areas. 

Let  P  be  the  pole  of  a  small 
circle  passing  through  the  three 
points,  ABO,  and  connect  P 
with  each  of  the  points,  A,  B, 

and  O,  by  arcs  of  great  circles.  Next,  through  E  draw 
the  arc  of  a  great  circle,  UP',  making  the  angle  DEP1 
equal  to  the  angle  ABP.  Take  EP'  =  BP,  and  draw 
the  arcs  of  great  circles,  P'D,  P'F. 

The  A's,  ABP  and  DEP',  are  equal  in  all  their  parts, 
because  AB=DE,  BP=EPf,  and  the  [_ABP=[_DEPf, 
(Prop.  9).  Taking  from  the  [_ ABO  the  [_ABP,  and 
from  the  [_DEF  the  [_DEPf,  we  have  the  remaining 
angles,  PBO  and  P'EF,  equal;  and  therefore  the  A's, 
BOP  and  EFP' ,  are  also  equal  in  all  their  parts. 

Now,  since  the  a's,  ABP  and  DEP',  are  isosceles,  they 
will  coincide  when  applied,  as  will  also  the  A's,  BOP 
and  EFP' ,  for  the  same  reason.  The  polygonal  areas, 
ABOP  and  DEEP',  are  therefore  equivalent.  If  from 
the  first  we  take  the  isosceles  triangle,  PAO,  and  from  the 
second  the  equal  isosceles  triangle,  P'DF,  the  remainders, 
or  the  triangles  ABO  and  DEF,  will  be  equivalent. 

Remark.  —  It  is  assumed  in  this  demonstration  that  the  pole  P  falls 
without  the  triangle.  Were  it  to  fall  within,  instead  of  without,  no 
other  change  in  the  above  process  would  be  required  than  to  add  the 
isosceles  triangles,  PAO,  P'DF,  to  the  polygonal  areas,  to  get  the 
areas  of  the  triangles,  ABC,  DEF. 


SECTION    I.  325 

Cor.  Two  spherical  triangles  on  the  same  sphere,  or  on 
equal  spheres,  will  be  equivalent  —  1st,  when  they  are 
mutually  equilateral;  —  2d,  when  they  are  mutually  equi- 
angular ;  —  3d,  when  two  sides  of  the  one  are  equal  to 
two  sides  of  the  other,  each  to  each,  and  the  included 
angles  are  equal ;  —  4th,  when  two  angles  of  the  one  are 
equal  to  two  angles  of  the  other,  each  to  each,  and  the 
included  sides  are  equal. 

PROPOSITION    XIV. 

If  two  arcs  of  great  circles  intersect  each  other  on  the  sur- 
face of  a  hemisphere,  the  sum  of  either  two  of  the  opposite  tri- 
angles thus  formed  will  be  equivalent  to  a  lune  whose  angle  is 
the  corresponding  angle  formed  by  the  arcs. 

Let  the  great  circle,  AEBC,  be  the  base  of  a  hemi- 
sphere, on  the  surface  of  which  the  semi-great  circumfer- 
ences, BBA  and  CBE,  inter- 
sect each  other  at  B ;  then  will 
the  sum  of  the  opposite  tri- 
angles, BBC  and  BAB,  be 
equivalent  to  the  lune  whose 
angle  is  BBC;  and  the  sum 
of  the  opposite  triangles, 
CB A  and  BBE,  will  be  equiv- 
alent to  the  lune  whose  angle 
is  CBA. 

Produce  the  arcs,  BBA  and 
CBE,  until  they  intersect  on  the  opposite  hemisphere  at  H; 
then,  since  CBB  and  BEH  are  both  semi-circumferences 
of  a  great  circle,  they  are  equal.  Taking  from  each  the 
common  part  BE,  we  have  CB  =HE.  In  the  same  way 
we  prove  BB  =  HA,  and  AE  =  BC.  The  two  triangles, 
BBC  and  HAE,  are  therefore  mutually  equilateral,  and 
hence  they  are  equivalent,  (Prop.  13).  But  the  two  tri- 
angles, HAE  and  ABE,  together,  make  up  the  lune 
28 


326 


SPHERICAL    GEOMETRY. 


DEHAD-,  hence  the  sum  of  the  a's,  B DO  and  ADE,  is 
equivalent  to  the  same  lune. 

By  the  same  course  of  reasoning,  we  prove  that  the 
sum  of  the  opposite  A's,  DAO  and  DBJE,  is  equivalent 
to  the  lune  BOH  AD,  whose  angle  is  ABC. 


PROPOSITION   XV. 

The  surface  of  a  lune  is  to  the  whole  surface  of  the  sphere, 
as  the  angle  of  the  lune  is  to  four  right  arigles ;  or,  as  the  arc 
which  measures  that  angle  is  to  the  circumference  of  a  great 
circle. 

Let  ABFOA  be  a  lune  on  the 
surface  of  a  sphere,  and  BOB 
an  arc  of  a  great  circle,  whose 
poles  are  A  and  F,  the  vertices 
of  the  angles  of  the  lune.  The 
arc,  BO,  will  then  measure  the 
angles  of  the  lune.  Take  any 
arc,  as  BD,  that  will  be  con- 
tained an  exact  number  of  times 
in  BO,  and  in  the  whole  circum- 
ference, BOJEB,  and,  beginning  at  B,  divide  the  arc  and 
the  circumference  into  parts  equal  to  BB,  and  join  the 
points  of  division  and  the  poles,  by  arcs  of  great  circles. 
We  shall  thus  divide  the  whole  surface  of  the  sphere 
into  a  number  of  equal  lunes.  Now,  if  the  arc  BO  con- 
tains the  arc  BB  m  times,  and  the  whole  circumference 
contains  this  arc  n  times,  the  surface  of  the  lune  will 
contain  m  of  these  partial  lunes,  and  the  surface  of  the 
sphere  will  contain  n  of  the  same ;  and  we  shall  have, 
Surf,  lune  :  surf,  sphere  : :  m  :  n. 

But,     m  :  n  : :  BO  :  circumference  great  circle ;   • 
hence,  surf,  lune  :  surf,  sphere  : :  BO  :  cir.  great  circle; 
or,         surf,  lune  :  surf,  sphere  ::  [__BOO :  4  right  angles. 


SECTION    I.  327 

This  demonstration  assumes  that  BD  is  a  common 
measure  of  the  arc,  BC,  and  the  whole  circumference.  It 
may  happen  that  no  finite  common  measure  can  be 
found ;  but  our  reasoning  would  remain  the  same,  even 
though  this  common  measure  were  to  become  indefinitely 
small. 

Hence  the  proposition. 

Cor.  1.  Any  two  lunes  on  the  same  sphere,  or  on  equal 
spheres,  are  to  each  as  their  respective  angles. 

Scholium.  —  Spherical  triangles,  formed  by  joining  the  pole  of  an 
arc  of  a  great  circle  with  the  extremities  of  this  arc  by  the  arcs  of 
great  circles,  are  isosceles,  and  contain  two  right  angles.  For  this 
reason  they  are  called  bi-redangular.  If  the  base  is  also  a  quadrant, 
the  vertex  of  either  angle  becomes  the  pole  of  the  opposite  side,  and 
each  angle  is  measured  by  its  opposite  side.  The  three  angles  are  then 
right  angles,  and  the  triangle  is  for  this  reason  called  tri-rectangular. 
It  is  evident  that  the  surface  of  a  sphere  contains  eight  of  its  tri- 
rectangular  triangles. 

Car.  2.  Taking  the  right  angle  as  the  unit  of  angles, 
and  denoting  the  angle  of  a  lune  by  A,  and  the  surface 
of  a  tri-rectangular  triangle  by  T,  we  have, 

surf,  of  lune  :  ST  ::  A  :  4; 
whence,       surf,  of  lune  =  2 A  x  T. 

Cor.  3.  A  spherical  ungula  bears  the  same  relation  to 
the  entire  sphere,  that  the  lune,  which  is  the  base  of  the 
ungula,  bears  to  the  surface  of  the  sphere ;  and  hence, 
any  two  spherical  ungulas  in  the  same  sphere,  or  in 
equal  spheres,  are  to  each  other  as  the  angles  of  their  re- 
spective lunes. 

PROPOSITION    XVI. 

The  area  of  a  spherical  triangle  is  measured  by  the  excess 
of  the  sum  of  its  angles  over  two  right  angles,  multiplied  by 
the  tri-rectangular  triangle. 

Let  AB C  be  a  spherical  triangle,  and  DEFLK  the  cir- 
cumference of  the  base  of  the  hemisphere  on  which  this 
triangle  is  situated. 


328  SPHERICAL    GEOMETRY. 

Produce  the  sides  of  tlie  tri- 
angle until  they  meet  this  cir- 
cumference in  the  points,  D,  U, 
F,  L,  K,  and  P,  thus  forming 
the  sets  of  opposite  triangles, 
FAF,  AKL ;  BFF,  BFK;  OFF, 
OFF. 

Now,  the  triangles  of  each  of 
these  sets  are  together  equal  to 
a  lune,  whose  angle  is  the  cor- 
responding angle  of  the  triangle,  (Prop.  14) ;  hence  we 
have, 

A  FAF  +  A  AKL  =  2 A  x  T,  (Prop. 15,  Cor.  2). 

ABFF  +  ABFK=2B  x  T 

A  OFF  +  A  CDF  =  2(7  x  T 

If  the  first  members  of  these  equations  be  added,  it  is 
evident  that  their  sum  will  exceed  the  surface  of  the 
hemisphere  by  twice  the  triangle  ABO;  hence,  adding 
these  equations  member  to  member,  and  substituting  for 
the  first  member  of  the  result  its  value,  4T  -f  2  A  ABO, 
we  have 

4T  +  2aAB0  =  2A.T  -f  2B.T  +  20.T 

or,      2T  +    AABO=   A.T  +    B.T  +    O.T 

whence,  A  ABO  =    A.T  +    B.T  +    O.T—2T. 

That  is,       AABO  =  (A  -f  B  +  0—  2)  T. 

But  A  -f  B  +  (7 —  2  is  the  excess  of  the  sum  of  the 
angles  of  the  triangle  over  two  right  angles,  and  T  de- 
notes the  area  of  a  tri-rectangular  triangle. 

Hence  the  proposition  ;  the  area,  etc. 


SECTION    I.  329 

PROPOSITION    XVII. 

The  area  of  any  spherical  polygon  is  measured  by  the  excess 
of  the  sum  of  all  its  angles  over  two  right  angles,  taken  as 
many  times,  less  two,  as  the  polygon  has  sides,  multiplied  by 
the  tri-rectangular  triangle. 

Let  AB  CJDE  be  a  spherical  poly-  £ 

gon;  then  will  its  area  be  meas-  j\^^*^      I 

ured  by  the  excess  of  the  sum  of  /\  / 

the  angles,  A,  B,  0,  D,  and  E,  over  / 

two  right  angles  taken  a  number         /  N.x 

of  times  which  is  two   less  than     c/ """" ""^>E 

the  number  of  sides,  multiplied  by        \  z' 

T,   the    tri  -  rectangular    triangle.  \.      / 

Through  the  vertex  of  any  of  the  ^ 

angles,  as  E,  and  the  vertices  of 

the  opposite  angles,  pass  arcs  of  great  circles,  thus  divi- 
ding the  polygon  into  as  many  triangles,  less  two,  as  the 
polygon  has  sides.  The  sum  of  the  angles  of  the  several 
triangles  will  be  equal  to  the  sum  of  the  angles  of  the 
polygon. 

Now,  the  area  of  each  triangle  is  measured  by  the 
excess  of  the  sum  of  its  angles  over  two  right  angles, 
multiplied  by  the  tri-rectangular  triangle.  Hence  the 
sum  of  the  areas  of  all  the  triangles,  or  the  area  of  the 
polygon,  is  measured  by  the  excess  of  the  sum  of  all  the 
angles  of  the  triangles  over  two  right  angles,  taken  as 
many  times  as  there  are  triangles,  multiplied  by  the  tri- 
rectangular  triangle.  But  there  are  as  many  triangles  as 
the  polygon  has  sides,  less  two. 

Hence  the  proposition ;  the  area  of  any  spherical  poly- 
gon, etc. 

Cor.  If  S  denote  the  sum  of  the  angles  of  any  spherical 
polygon,  n  the  number  of  sides,  and  T  the  tri-rectan- 
gular triangle,  the  right  angle  being  the  unit  of  angles ; 
the  area  of  the  polygon  will  be  expressed  by 

IS—  2  (w-2)]x  T=  (S—  2n  +  4)  T. 
28* 


330 


SPHERICAL    TRIGONOMETRY. 


SECTION  II. 


SPHERICAL    TRIGONOMETRY. 

A  Spherical  Triangle  contains  six  parts — three  sides  and 
three  angles — any  three  of  which  being  given,  the  other 
three  may  be  determined. 

Spherical  Trigonometry  has  for  its  object  to  explain  the 
different  methods  of  computing  three  of  the  six  parts  of 
a  spherical  triangle,  when  the  other  three  are  given.  It 
may  be  divided  into  Right-angled  Spherical  Trigonome- 
try, and  Oblique-angled  Spherical  Trigonometry ;  the  first 
treating  of  the  solution  of  right-angled,  and  the  second 
of  oblique-angled  spherical  triangles. 

RIGHT-ANGLED   SPHERICAL  TRIGONOMETRY. 


PROPOSITION   I. 

With  the  sines  of  the  sides,  and  the  tangent  of  ONE  SIDE 
of  any  right-angled  spherical  triangle,  two  plane  triangles  can 
be  formed  that  will  be  similar,  and  similarly  situated. 

Let  ABO  be  a  spherical  triangle, 
right-angled  at  B ;  and  let  D  be  the 
center  of  the  sphere.  Because  the 
angle  OB  A  is  a  right  angle,  the  plane 
OBD  is  perpendicular  to  the  plane 
DBA.  From  0  let  fall  OR,  perpen- 
dicular to  the  plane  DBA ;  and  as  the 


SECTION    II.  331 

plane  CBD  is  perpendicular  to  the  plane  DBA,  CE  will 
lie  in  the  plane  OBI),  and  be  perpendicular  to  the  line 
DB,  and  perpendicular  to  all  lines  that  can  be  drawn  in 
the  plane  DBA,  from  the  point  E  (Def.  2,  B.  VI). 

Draw  EG  perpendicular  to  DA,  and  draw  GrC;  GC 
will  lie  wholly  in  the  plane  CDA,  and  CEG  is  a  right- 
angled  triangle,  right-angled  at  E. 

We  will  now  demonstrate  that  the  angle  DGfC  is  a 
right  angle. 

The  right-angled  ACEG,  gives  CE2+EG2  =  CG2   fel) 
The  right-angled  AD GE,  gives  DG2+EG2=DE2   (2) 
By  subtraction,     CH2  —  DG2  =  CG2  —  DE2       ( 3 ) 
By  transposition,  OH2  +  DH2  =  CG2  +  DG2       (4) 

But  the  first  member  of  equation  (4)?  is  equal  to 
CD2,  because  ODE  is  a  right-angled  triangle; 

Therefore,  CD2  =  CG2  +  DG2 

Hence,  CD  is  the  hypotenuse  of  the  right-angled  tri- 
angle DGC,  (Th.  39,  B.  I). 

,  From  the  point  B,  draw  BE  at  right  angles  to  DA, 
and  BF  at  right  angles  to  DB,  in  the  plane  CDB  ex- 
tended ;  the  point  F  will  be  in  the  line  DC.  Draw  EF, 
and  as  F  is  in  the  plane  CDA,  and  i?  is  in  the  same 
plane,  the  line  EF  is  in  the  plane  CDA.  Now  we  are  to 
prove  that  the  triangle  CEG  is  similar  to  the  triangle 
BEF,  and  similarly  situated. 

As  EG  and  BE  are  both  at  right  angles  to  DA,  they 
are  parallel ;  and  as  EC  and  BF  are  both  at  right  angles 
to  DB,  they  are  parallel ;  and  by  reason  of  the  parallels, 
the  angles  GEC  and  EBF  are  equal ;  but  GEC  is  a  right 
angle ;  therefore,  EBF  is  also  a  right  angle. 

Now,  as  GE  and  BE  are  parallel,  and  CE  and  BF 
are  also  parallel,  we  have, 

DE  i  DB  =  EG:  BE 
And,  DE  :  DB  =EC  :  BF 


332  SPHERICAL   TRIGONOMETRY. 

Therefore,         HO  :  BE  =  HO  :  BF  (Th.  6,  B.  II), 
Or,  Ha  :  HO  =  BE  :  BF. 

Here,  then,  are  two  triangles,  having  an  angle  in  the 
one  equal  to  an  angle  in  the  other,  and  the  sides  about 
the  equal  angles  proportional;  the  two  triangles  are 
therefore  equiangular,  (Cor.  2,  Th.  17,  B.  II);  and  they 
are  similarly  situated,  for  their  sides  make  equal  angles 
at  H  and  B  with  the  same  line,  DB. 

Hence  the  proposition. 

Scholium.  —  By  the  definition  of  sines,  cosines,  and  tangents,  we 
perceive  that  CH  is  the  sine  of  the  arc  BC,  DH  is  its  cosine,  and  BF 
its  tangent;  CG  is  the  sine  of  the  arc  AC,  and  DG  its  cosine.  Also, 
BE  is  the  sine  of  the  arc  AB,  and  BE  is  the  cosine  of  the  same  arc. 
With  this  figure  we  are  prepared  to  demonstrate  the  following  propo- 
sitions. 

PROPOSITION    II. 

In  any  right-angled  spherical  triangle,  the  sine  of  one  side 
is  to  the  tangent  of  the  other  side,  as  radius  is  to  the  tangent 
of  the  angle  adjacent  to  the  first-mentioned  side. 

Or,  the  sine  of  one  side  is  to  the  tangent  of  the  other  side, 
as  the  cotangent  of  the  angle  adjacent  to  the  first-mentioned 
side  is  to  the  radius. 

For  the  sake  of  brevity,  we  will  represent  the  angles 
of  the  triangle  by  A,  B,  0,  and  the  sides  or  arcs  opposite 
to  these  angles,  by  a,  b,  c,  that  is,  a  opposite  A,  etc. 

In  the  right-angled  plane  triangle  EBF,  we  have, 
EB  :  BF  =  B  :  tzn.BEF 

That  is,     sin.c  :  tan.a  =  R  :  tan.J., 

which  agrees  with  the  first  part  of  the  enunciation.  By 
reference  to  equation  (5),  Section  I,  Plane  Trigonometry, 
we  shall  find  that, 

tan.JL  cot. A  =  B2; 

B2 

therefore,  tan.  J.  = -. 

cot.  A 


SECTION  II.  333 

Substituting  this  value  for  tangent  A,  in  the  preceding 
proportion,  and  dividing  the  last  couplet  by  R,  we  shall 
have, 

sin.c  :  tan.a  =  1  :  -. 

cot.JL 

Or,  sin.c  :  tan.a  =  cot. J.  :  R. 

Or,  R  sin.c  =  tan.a  cot. JL,  (1) 

which  answers  to  the  second  part  of  the  enunciation. 

Cor.  By  changing  the  construction,  drawing  the  tan- 
gent to  AB,  in  place  of  the  tangent  to  BC,  and  proceed- 
ing in  a  similar  manner,  we  have, 

R  sin. a  =  tan.c  cot.  C.  ( 2 ) 


PROPOSITION   III. 

In  any  right-angled  spherical  triangle,  the  sine  of  the  right 
angle  is  to  the  sine  of  the  hypotenuse,  as  the  sine  of  either  of 
the  other  angles  is  to  the  sine  of  the  side  opposite  to  that  angle. 

,    The  sine  of  90°,  or  radius,  is  designated  by  R. 
In  the  plane  triangle,  CMGr,  we  have, 

Bm.CMG-  :  Ca  =  sin.CG-H  :  CM 
That  is,  R  :  sin.6  =  sin.  J.  :  sin.a 

Or,  R  sin. a  =  sin. b  sin. A  ( 3 ) 

Cor.   By  a  change  in  the  construction  of  the  figure, 

drawing  a  tangent  to  AB,  etc.,  we  shall  have, 

R  :  sin.6  =  sin.<7  :  sin.c 
Or,  R  sin.c  =  sin.5  sin.  C.  ( 4 ) 

Scholium.  —  Collecting  the  four  equations  taken  from  this  and  the 
preceding  proposition,  we  have, 

( 1 )  iJTsin.c  =  tan.a  cot.  J. 

( 2 )  R  sin. a  =  tan.c  cot.  C 
( 3  )  B  sin.a  =  sin.6  sin.  A 
( 4 )  E  sin.c  =  sin. b  sin.  C 


334 


SPHERICAL    TRIGONOMETRY. 


These  equations  refer  to  the  right-angled 
triangle,  ABC;  but  the  principles  are  true 
for  any  right-angled  spherical  triangle.  Let 
us  apply  them  to  the  right-angled  triangle, 
PDC,  the  com  piemen  tal  triangle  to  ABC. 

Making  this  application,  equation 

( 1 )  becomes  R  sin.  CD  =  tan.PD  cot.  C 

( 2 )  becomes  R  a'm.PD  =  tan.  CD  cot.P 
( 3  )  becomes  R  sin. PD  =  sin. PC  sin.  C 
( 4 )  becomes  R  sin.  CD  =  ain.PC  sin.P 

By  observing  that  sin.  CD  =  cos.  J.  C  =  cos.6. 

And  that  tan.PD  =  cot.DO  =  cot. J.,  etc.;  and  by  running  equa- 
tions (  n  ),  (m ),  (  o  ),  and  (p  ),  back  into  the  triangle,  ABC,  we  shall 
have, 

( 5  )  R  cos. 6  =  cot.  J.  cot.  C 
(  6  )  R  cos. A  =  cot.6    tan.c 
(7)  R cos. A  =  cos.a  sin. C 
(  8  )  R  cos.b  —--  cos.a  cos.c 

By  observing  equation  (  6  ),  we  find  that  the  second  member  refers 
to  sides  adjacent  to  the  angle  A.  The  same  relation  holds  in  respect 
to  the  angle  C,  and  gives, 

(9)  Rco9.C=  cot.6  tan.a. 
Making  the  same  observations  on  ( 7 ),  we  infer, 
(10)  R  cos.  C  =  cos.c  sin.X 

Observation  1.  Several  of  these  equations  can  be  de- 
duced geometrically  without  the  least  difficulty.  For 
example,  take  the  figure  to  Proposition  1.  The  parallels 
in  the  plane,  DBA,  give, 

DB  :  VH=  DU  :  Da. 

That  is,  R  :  cos.a  =  cos.c  :  cos.5. 

A  result  identical  with  equation  ( 8 )?  and  in  words  it  is 
expressed  thus :  Radius  is  to  cosine  of  one  side,  as  the  cosine 
of  the  other  side  is  to  the  cosine  of  tiie  hypotenuse. 

Observation  2.  The  equations  numbered  from  (1)  to 
(10)  cover  every  possible  case  that  can  occur  in  right- 
angled  spherical  trigonometry ;  but  the  combinations  are 


SECTION   II. 


335 


too  various  to  be  remembered,  and  readily  applied  to  prac- 
tical use. 

"We  can  remedy  this  inconvenience,  by  taking  the  com- 
plement of  the  hypotenuse,  and  the  complements  of  the  two 
oblique  angles,  in  place  of  the  arcs  themselves. 

Thus,  b  is  the  hypotenuse,  and  let  V  be  its  complement. 

Then,  5  +  ^=90°;  or,  b=  90°  —  6';  and,  sin.6  =  cos.6', 
cos.6  =  sin. V ;  tan.5  =  cot.5r.  In  the  same  manner,  if  A1 
is  the  complement  to  A, 

Then,  sin. A  =  cos. J/;  cos. A  =  sin. J/;  and,  tan. J.  = 
cot. J/;  and  similarly,  sin. C—  cos. C;  cos. Q—  sin.C;  and 
tan.  (7=  cot.  C. 

Substituting  these  values  for  b,  A,  and  0,  in  the  fore- 
going ten  equations  (a  and  c  remaining  the  same),  we 
have, 


(11) 
(12) 
(13) 
(14) 
(15) 
(16) 
(17) 
(18) 
(19) 
(20) 


NAPIE 

Rsin.c  = 
Rsm.a  = 
Rsin.a  = 
R  sin.<?  = 
R  sin.6'  = 
_Ksin.^'= 
Rsm.Af= 
Rsm.b'  = 
Rsin.Cf= 
R8in.Cf= 


r's   circular 

=  tan.a  tan.  J/ 
=  tan.<?  tan.  (7 
■  cos.5'  cos.  J.' 
=  cos.5'  cos.  C 
=  tan. J/ tan. C" 
=  tan.5'  tan.c 
=  cos.a  cos.C" 
=  cos.a  cos.c 
=  tan. V  tan. a 
--  cos.c?  cos. A' 


PARTS. 

Omitting  the  consid- 
eration of  the  right  an- 
gle, there  are  five  parts. 
Each  part  taken  as  a 
middle  part,  is  connect- 
ed to  its  adjacent  parts 
by  one  equation,  and 
to  its  extreme  parts  by 
another  equation ;  there- 
fore, ten  equations  are 
required  for  the  combi- 
nations of  all  the  parts. 


These  equations  are  very  remarkable,  because  the  first 
members  are  all  composed  of  radius  into  some  sine,  and 
the  second  members  are  all  composed  of  the  product  of 
two  tangents,  or  two  cosines. 

To  condense  these  equations  into  words,  for  the  pur- 
pose of  assisting  the  memory,  we  will  refer  any  one  of 
them  directly  to  the  right-angled  triangle,  ABC,  in  the 
last  figure. 


336  SPHERICAL    TRIGONOMETRY. 

When  the  right  angle  is  left  out  of  the  question,  a 
right-angled  triangle  consists  of  five  parts  —  three  sides, 
and  two  angles.  Let  any  one  of  these  parts  be  called  a 
middle  part;  then  two  other  parts  will  lie  adjacent  to  this 
part,  and  two  opposite  to  it,  that  is,  separated  from  it  by 
two  other  parts. 

Tor  instance,  take  equation  ( 11 ),  and  call  c  the  middle 
part;  then  A1  and  a  will  be  adjacent  parts,  and  O  and  V 
opposite  parts.  Again,  take  a  as  a  middle  part;  then  c 
and  Q'  will  be  adjacent  parts,  and  A1  and  V  will  be  oppo- 
site parts ;  and  thus  we  may  go  round  the  triangle. 

Take  any  equation  from  ( 11 )  to  (20 )?  and  consider  the 
middle  part  in  the  first  member  of  the  equation,  and  we 
shall  find  that  they  correspond  to  the  two  following  inva- 
riable and  comprehensive  rules  : 

1.  The  radius  into  the  sine  of  the  middle  part  is  equal  to 
the  product  of  the  tangents  of  the  adjacent  parts. 

2.  The  radius  into  the  sine  of  the  middle  part  is  equal  to 
the  product  of  the  cosines  of  the  opposite  parts. 

These  rules  are  known  as  .Napier's  Rules,  because  they 
were  first  given  by  that  distinguished  mathematician, 
who  was  also  the  inventor  of  logarithms. 

In  the  application  of  these  equations,  the  accent  may  be 
omitted  if  tan.  be  changed  to  cotan.,  sin.  to  cosin.,  etc. 
Thus,  if  equation  ( 13 )  were  to  be  employed,  it  would  be 
written,  in  the  first  instance,  Msm.a  =  cos.6'  cos.  J/,  to 
insure  conformity  to  the  rule ;  then,  we  would  change  it 
into  B  sin.a  =  sin.5  sin. A. 

Remark. — "We  caution  the  pupil  to  be  very  particular  to  take  the 
complements  of  the  hypotenuse,  and  the  complements  of  the  oblique 
angles. 


SECTION    III. 


337 


SECTION   III 


OBLIQUE-ANGLED   SPHERICAL   TRIGONOMETRY. 

The  preceding  investigations  have  had  reference  to 
right-angled  spherical  trigonometry  only,  but  the  appli- 
cation of  these  principles  cover  oblique-angled  trigonom- 
etry also;  for,  every  oblique-angled  spherical  triangle 
may  be  considered  as  made  up  of  the  sum  or  difference 
of  two  right-angled  spherical  triangles.  With  this  ex- 
planatory remark,  we  give 


PROPOSITION   I. 

In  all  spherical  triangles,  the  sines  of  the  sides  are  to  each 
other,  as  the  sines  of  the  angles  opposite  to  them. 

This  was  proved  in  relation  to  right-angled^  triangles  in 
Prop.  3,  Sec.  II,  and  we  now  apply  the  principle  to  ob- 
lique-angled triangles. 

Let  ABO  be  the  triangle,  and  let 
CD  be  perpendicular  to  AB,  or  to 
AB  produced. 

Then,  by  Prop.  3,  Sec.  II,  we  have, 

B  :  sin.  A  0  =  sin.  A  :  sin.  OB. 
Also, 
sin.  OB  :  R  =  sin.CZ)  :  sin.  B. 
29  w 


338  SPHEKICAL    TRIGONOMETRY. 

By  multiplying  these  two  proportions  together,  term 
by  term,  and  omitting  the  common  factor  B,  in  the  first 
couplet,  and  the  common  factor,  sm.CD,  in  the  second, 
we  have 

sin. CB  :  sin.  AC  =  sin. J.  :  sin.B. 


PROPOSITION  II. 

In  any  spherical  triangle,  if  an  arc  of  a  great  circle  be  let 
fall  from  any  angle  perpendicular  to  the  opposite  side  as  a 
base,  or  to  the  base  produced,  the  cosines  of  the  other  two 
sides  will  be  to  each  other  as  the  cosines  of  the  segments  of 
the  base. 

By  the  application  of  equation  8,  (Sec.  II),  to  the  last 
figure,  we  have, 

R  cos. A  C  mm  cos. AD  eos.DC 
Similarly,      B  cos.BC  ==  eos.DC  cos.BD 
Dividing  one  of  these  equations  by  the  other,  omitting 
common  factors  in  numerators   and  denominators,  we 
have, 

cos. A  C  _    cos. AD 

cos.BC        cos.BD 
Or,        cos. A C  :  cos.BC  =  cos. AD  :  coa.BD. 

PROPOSITION   III. 

If  from  any  angle  of  a  spherical  triangle,  a  perpendicular 
be  let  fall  on  the  base,  or  on  the  base  produced,  the  tangents 
of  the  segments  of  the  base  will  be  reciprocally  proportional 
to  the  cotangents  of  the  segments  of  the  angle. 

By  the  application  of  Equation  2,  (Sec.  II),  to  the  last 
figure,  we  have, 

B  sin.CD  =  tan. AD  cot. ACD. 


SECTION   III.  339 

Similarly,     R  Bin.  CD  =  t&n.BD  cot.BOD 
Therefore,  by  equality, 

t&n.AD  cot.ACD  =  tan.BD  cot.BOD 
Or,       tan. AD  :  t&n.BD  =  cot.BOD  :  cot. AOD. 

PROPOSITION   IV. 

The  same  construction  remaining,  the  cosines  of  the  angles 
at  the  extremities  of  the  segments  of  the  base  are  to  each 
other  as  the  sines  of  the  segments  of  the  opposite  angle. 
Equation  7,  (Sec.  II),  applied  to  the  triangle  AOD,  gives 

R  cob. A  =  cos. OD  Bin. AOD      (s) 
Also,         R  cob.B  »  cob.OD  Bin.BOD      (t) 
Dividing  equation  (*)  by  (0,  gives 
cos. A  _  sin. AOD 
cob.B       Bin.BOD 
Or,  cos.i?  :  cos. A  =  Bin.BOD  :  sin. AOD. 

PROPOSITION   V. 

The  same  construction  remaining,  the  sines  of  the  segments 
of  the  base  are  to  each  other  as  the  cotangents  of  the  adjacent 
angles. 
Equation  1,  (Sec.  II),  applied  to  the  triangle  AOD,  gives 

R  Bin. AD  -  tan.  OD  cot.  A  ( * ) 
Similarly,  R  Bm.BD  =  t&n.OD  cot.B  (J) 
Dividing  («)  by  (0,  gives 

Bin.AD  _  cot.A 
Bin.BD       cot.B 
Or,  Bin.BD  :  sin. AD  =  eot.i?  :  cot.A. 


340 


SPHERICAL    TRIGONOMETRY. 


PROPOSITION    VI. 

The  same  construction  remaining,  the  cotangents  of  the  two 
sides  are  to  each  other  as  the  cosines  of  the  segments  of  the 
angle. 
Equation  9,  (Sec.  II),  applied  to  the  triangle  A  CD,  gives 

B  cos.  ACD  =  eot.  AC  tun.  CD  (*) 
Similarly,  E  cos.BCD  =  cot.  BC  tan.  CD  (0 
Dividing  (*)  by  (0,  gives 

cos.  ACD       cot.  AC 


Or, 


cos.BCD       cot.BC 
cotAC:  cot.BC  =  cos.ACD  :  cos.BCD. 


PROPOSITION    VII. 

The  cosine  of  any  side  of  a  spherical  triangle,  is  equal  to 
the  product  of  the  cosines  of  the  other  two  sides,  plus  the 
product  of  the  sines  of  those  sides  multiplied  by  the  cosine 
of  the  included  angle. 

Let  ABC  be  a  spherical  triangle, 
and  CD  a  perpendicular  from  the 
angle  C  on  to  the  side  AB,  or  on  to 
the  side  AB  produced.  Then,  by 
Prop.  2, 
cos.  AC  :cos.CB=cos.  AD:  cos.BD{l) 

When  CD  falls  within  the  tri- 
angle, 

BD  =  (AB—  AD); 

and  when  CD  falls  without  the  triangle, 
BD  =  (AD—  AB). 

Hence,  cos.BD  —  cos.(AZ)  —  AB) 

Now,         cos.(AB  —  AD)  =  cos.(AD  —  AB), 

because  each  of  them  is  equal  to 
cos.AB  cos.AD  +   sin.AB  sin.AD,   (Eq.  10,   Prop.  2, 
Sec.  I,  Plane  Trig.). 


SECTION    III.  341 

This  value  of  cos.  BD,  put  in  proportion  ( 1 ),  gives 
cos.AC:  cos.  CB  =  cos.AD  :  cos.  AB  cos.AD+sinAB  sin.AD  (2) 

Dividing  the  last  couplet  of  proportion  ( 2 )  by  cos. AD, 

observing  that 

sin. AD        ,        .  -~ 
—  =  tan.J.D, 

COS.AD  ' 

and  we  have 

cos. A 0  :  cos. OB  ==  1  :  cos. AB  +  sin. AB  tan.  AD     (3) 

By  applying  equation  6,  (Sec.  II),  to  the  triangle  ACD, 
taking  the  radius  as  unity,  we  have 

cos.  J.  =  cot.AO  t&n.AD      (&) 

But,  tan.^4(7cot.^(7=-l,(Eq.  5,  Seal,  Plane  Trig.)     (0 

Multiply  equation  (&)  by  tan.  A O,  observing  equation 
(I),  and  we  have 

tan.  A  0  cos.  A  =  tan.  AD 

Substituting  this  value  of  tan.  AD,  in  proportion  (3), 
we  have 
cos. AC:  cos. OB  =  1 :  cos. AB  +  sin. AB  tan. AO cos. A  (4) 

Multiplying  extremes  and  means,  gives 
cos.  OB— cos.  A  0'  cos.  AB+ sin.  AB  (cos.  J.  0  tan.  A  0)  cos.  J.. 

But,     tan.JL<7= '--r^,  or,  cos. A 0  tan. A 0=  sin. AO. 

cos.  AO 

Therefore,  cos. OB  =  cos. AO  cos.AZ?+  sin.JJ?  sin. A 0 
cos.J.. 

If  the  sides  opposite  the  angles,  A,  B,  and  0,  be  re- 
spectively represented  by  a,  b,  and  e,  this  equation 
becomes, 

cos.a  =  cos.6  cos.c-l- sin.&  sin.e  cos.  J.. 

This  formula  conforms  to  the  enunciation  in  respect  to 
the  side  a.  Now,  by  simply  writing  b  for  a,  and  B  for  A, 
in  the  last  equation,  we  get  the  formula  for  cos. b,  which  is, 

cos.6  =  cos.a  cos.c  +  sin.a  sin.c  cos.B. 
29* 


342  SPHERICAL    TRIGONOMETRY. 

By  writing  c  for  a,  and  0  for  A,  we  get  the  formula  for 
cos.c,  which  is, 

cos.c  =  cos.a  cos.b  +  sin.a  sin.5  cos.C. 
Hence,  we  have  the  three  symmetrical  formulae : 

cos.a  =  cos.6  cos.c  -f  sin. b  sin.c  cos.A^ 
cos.6  =  cos.a  cos.c  +  sin. a  sin.c  cos.i?  >    \S) 
cos.c  =  cos.  a  cos.5  +  sin.a  sin.5  cos.C  J 

From  these,  by  simple  transposition  and  division,  we 
deduce  the  following  formulas  for  the  cosines  of  the 
angles  of  any  spherical  triangle,  viz : 

cos.a —  cos.5  cos.c^ 


cos.  A  = 
cos.i?  = 
cos.  0  = 


sin.&' sin.c 
cos. b  —  cos.a  cos.c 

sin. a  sin.c 
cos.c  —  cos.a  cos. b 

sin. a  sin. b 


(S>) 


By  means  of  these  equations  we  can  find  the  cosine  of 
any  of  the  three  angles  of  a  spherical  triangle  in  terms 
of  the  functions  of  the  sides ;  but  in  their  present  form 
they  are  not  suited  for  the  employment  of  logarithms, 
and  we  should  be  compelled  to  use  a  table  of  natural 
sines  and  cosines,  and  to  perform  tedious  numerical  ope- 
rations, to  obtain  the  value  of  the  angle. 

They  are,  however,  by  the  following  process,  trans- 
formed into  others  well  adapted  to  the  use  of  logarithms. 

In  Eq.  34,  Sec.  I,  Plane  Trig.,  we  have 

1  +  cos.  A  =  2cos.2J^4. 

m,         p  a        ,-,   a        -t    ,   cos.a — COS.6  cos.c 

Therefore,  2cos.2 \A  =  1  + : — ; — : •. 

sm.6  sin.c 

(sin. b  sin.c  —  cos. b  cos.c)  +  cos.a  ,    » 
sm.o  sin.c 

But,  cos.(6  +  c)  =  cos.5  cos.c  =  sin.c  sin. b,  (Equation 
9,  Section  I,  Plane  Trig.).     By  comparing  this  equation 


SECTION    III. 


343 


with  the  second  member  of  equation  ( »*  ),  we  perceive 
that  equation  ( m )  is  readily  reduced  to 

sin. b  sin.c 

Considering  (b+c)  as  one  are,  and  then  making  appli- 
cation of  equation  ( 18 ),  Plane  Trigonometry,  we  have, 


But, 


2cos.2|J.= 

b  -\-  c —  a      b  -f  c  +  a 
2  2 


W  (—,— )  ».n.  (—- 


') 


sin. 6  sin. c 
—  a;   and  if  we  put  S  to 


A  -J-  /»  _j_   /y 

represent — ,  we  shall  have, 

A  __  sin.#  sin.(# —  a) 


COS.'  -—  = 


Or, 


COS. 


2  sin.6  sin.c 

A  _  *    /sm.S  sm.(S — a) 
2        Y        sin. 6  sin.c 


'  The  second  member  of  this  equation  gives  the  value 
of  the  cosine  when  the  radius  is  unify.  To  a  greater 
radius,  the  cosine  would  be  greater;  and  in  just  the  same 
proportion  as  the  radius  increases,  all  the  trigonometrical 
lines  increase ;  therefore,  to  adapt  the  above  equation  to 
our  tables  where  the  radius  is  M,  we  must  write  R  in  the 
second  member,  as  a  factor;  and  if  we  put  it  under  the 
radical  sign,  we  must  write  TC\ 

For  the  other  angles  we  shall  have  precisely  similar 
equations : 

That  is,     cos.  -  =  \  /-fl2sin.ff  sin.(-#—  a) 


2         v  sin. b  sin.<? 

cos.  -  =  v/fl'sin-A  sin7(ff=^) 

2t  V  oin 


sin. a  sin. c 


cos 


■S-\/s 


sin.AS'  sin.(# — c) 


sin. a  sin. b 


(f) 


344  SPHERICAL    TRIGONOMETRY. 

To  deduce  from  formulae  (S),  formulae  for  the  sines  of 
the  half  of  each  of  the  angles  of  a  spherical  triangle,  we 
proceed  as  follows :. 

From  Eq.  35,  Sec.  I,  Plane  Trig.,  we  have 

2sin.2  J  J.  =  1  —  cos.  J.. 

Substituting  the  value  of  cos. A,  taken  from  formulae 
(S),  and  we  have, 

0  .    , .  A      .,      eos.a  —  cos.b  C08.c 

2sm.2  IA  =  1  — : — . — j . 

sin. 6  sin.c 

_  (sin.5  sin.e-f  cos.b  cos.c)  —  cos.a      ,   . 
sin.6  sin.c 

But,  cos.(5  co  c)  —  sin.5  sin.e.  -f  cos.5  cos.e,  (Eq.  10, 
Sec.  I,  Plane  Trig.). 

This  equation  reduces  equation  ( o )  to 

0.    .,  A      cos.(bcoc)  —  cos.a 
sm.o  sm.c 

Considering  (b  c*  c)  as  a  single  arc,  and  applying  equa- 
tion 18,  Sec.  I,  Plane  Trig.,  we  have 

~.      /a  4-  b  —  C\    .      /a  +  o  —  b\ 
2sin.  ( )  sin.  ( - ) 

Mn.'M-'        ^       2-    :    •  2         •  ^ 

sm.6  sin.c 

-r»   ,    «  +  & —  e      a  +  b  +  c  a  .»  ,    a 

But, —  =  — -- c  =  aS" —  <?,  if  we  put  S  = 

A  A 

a  +  b  +  e 

2~~* 

A1       a  +  c—b      a+b-hc      ,       ~      , 
Also, = - b  =  S—b. 

Dividing  equation  ( o' )  by  2,  and  making  these  substi- 
tutions, we  have 

sin  2  M  •  -  ^'^  ~  c)  sin'^  ~~  J) 

Bin.     2^- : 7 ? > 

sm.6  sin.c 
when  radius  is  unity. 


SECTION    III 
When  radius  is  B,  we  have 

v  sin. b  sin.c 

Similarly,  sin.J*  =  ^H^EfeLtl) 
v  sin.fl  sin.c 

.     ;  „       v  /S2sin.(iS'  — «)sin.(^— 6 
sin.J(7  =  y  — 


345 


And, 


sin. a  sin.6 


IV) 


To  apply  to  our  tables,  B2  must  be  put  under  the  radi- 
cal sign.  We  shall  show  the  application  of  these  form- 
ulae, and  those  in  group  (T),  hereafter. 


PROPOSITION    VIII. 

The  cosine  of  any  of  the  angles  of  a  spherical  triangle,  is 
equal  to  the  product  of  the  sines  of  the  other  two  angles  mul- 
tiplied by  the  cosine  of  the  included  side,  minus  the  product 
of  the  cosines  of  these  other  two  angles. 

•  Let  ABO  be  a  spherical  triangle,  and 
ArBrQr  its  supplemental  or  polar  tri- 
angle, the  angles  of  the  first  being  de- 
noted by  A,  B,  and  0,  and  the  sides 
opposite  these  angles  by  a,  5,  c,  respect- 
ively ;  A',  Bf,  Qr,  a',  V ,  cf,  denoting  the 
angles  and  corresponding  sides  of  the 
second. 

By  Prop.  5,  Sec.  I,  we  have  the  following  relations  be- 
tween the  sides  and  angles  of  these  two  triangles. 

A'  m  180°  —  a,Bf  =  180°  —  b,  C  =  180°  —  c; 
a'  -  180°  —  A,V=  180°  —  B,  c1  =  180°  —  C. 

The  first  of  formulae  (#),  Prop.  7,  when  applied  to  the 
polar  triangle,  gives 

cos. a'  =  cos.5'  cos.c'  +  sin.5;  sin.c'  cos. Ar     (1) 


346  SPHERICAL    TRIGONOMETRY. 

which,  by  substituting  the  values  of  a',  bf,  c',  and  A\ 
becomes 

cos.(180° — A)  =  cos.(180° — B)  cos.(180°  —  C)  +  sin.(180° 
—  B)  sin.(180°  —  0)  cos.(180°  —  a),      ( 2 ) 

But, 

cos.(180°— A)  =  —  cos.^4,  etc.,  sin.(180°— B)  =  sin.B,  etc. ; 

and  placing  these  values  for  their  equals  in  eq.  ( 2 ),  and 
changing  the  sines  of  both  members  of  the  resulting 
equation,  we  get 

cos.  J.  =  sin.i?  sm.O  cos.a —  cos.i?  cos.  O, 

which  agrees  with  the  enunciation. 

By  treating  the  other  two  of  .formulae  (£),  Prop.  7,  in 
the  same  manner,  we  would  obtain  similar  values  for  the 
cosines  of  the  other  two  angles  of  the  triangle  ABC; 
or  we  may  get  them  more  easily  by  a  simple  permuta- 
tion of  the  letters  A,  B,  C,  a,  etc. 

Hence,  we  have  the  three  equations 

cos. A  =  sin.i?  sin.  (7  cos.a  —  cos. B  cos.  C^ 
C08.B  mm  sin.  A  sm.O  cos. b  —  cos. A  cos.  0  V     CO 
cos.  Q  =  sin.-d  sin.i?  cos.c?  —  cos. A  cos.B  ) 

By  transposition  and  division,  these  equations  become 

cos. A  -f  cos.B  cos. 0    ,o\ 
cos. a  = {o) 

sm.B  sin.  C 

,       cos.B  4-  cos. A  cos.  0 

sin. A  sm.C7 

cos.  0  +  cos. A  cos.B 

cos.c  = r— — , — i_ 

S111..A  sinJ 

From  these  we  can  find  formulae  to  express  the  sine  or 
the  cosine  of  one  half  of  the  side  of  a  spherical  triangle, 
in  terms  of  the  functions  of  its  angles ;  thus : 

Add  1  to  each  member  of  eq.  (3),  and  we  have 

cos.  J.  +  cos.i?  cos.  O  -f  sin.  B  sin.  0 


1  +  cos.a  = 


sin.i?  sin.6Y 


SECTION    III.  347 

cos.  A  +  cos.(J5  —  0) 
sin.2?  sin.G7 

But,  1  +  cos.a  =  2cos.2  \a ;  hence, 

0       , ,         cos.^L  +  cos.fi?  —  C) 

2cos.2i«  - 4    tf  }>„ i 

sin.J?  sin.  (7 

and  since  cos. J.  4-  cos.(2?  —  C)  =  2cos.J(J.  +  i? —  (7)cos.J 
(4+a—JB)  (Eq.17,  Sec.  I,  Plane  Trig.),  we  have 

2cos.2  la  =  2cos-*(^  +  B-C)coB.i(A  +  (7-i?) 
2  sin.J5  sin.67 

Make  A  +  B  +  0=2S;  then  A  +  B—C=2S—2C, 
A+C—B  =  2S—2B,  i(A  +  B—C)  =  jS—C,an&  ±{A 
+  C—B)  =  S—B\  whence 

sm.jo  sm.(7 

4  /cos.^— <7)cos.(/SCTg5 
or,  cos.fa  =  \/ ^— = — £? .  >  ^ c 

1  v  sm.2?sin.<7 

,     Similarly,  cos.|5  =  V       ^/^ -;  / 

and,  cos.Jc  -  V^S^E! 

v  sin. J.  sinJ 

.To  find  the  sin.Ja  in  terms  of  the  functions  of  the 
angles,  we  must  subtract  each  member  of  eq.  ( 3 )  from  1, 
by  which  we  get 

H  _,      cos.  J.  +  cos.i?  cos.  (7 

1  —  cos.a  =  l r  .        . 

sui.jd  sm.G 

But,  1 — cos.a  =  2sin.2  Ja ;  hence  we  have, 

o-    2i    _(sin.J5  sin. (7 — cos. 5  cos. (7) — cos.JL 
smJ  sin.CT 

Operating  upon  this  in  a  manner  analogous  to  that  by 
which  cos.Ja  was  found,  we  get, 


348  SPHERICAL   TRIGONOMETRY. 

t         smJ  sm.  0        J 

.    l4       f— cos.#  cos.(S-B)  l  $ 

sin. lb  =  \ . — —A-jy — J-  V 2       )     (W) 

\         sin.  J.  sin.  (7        J 

.     i  f  —  cos.#cos.(#—  O)}  i 

Sin. Jc  =  ■{  : — r- L_- '   I  2 

t        sin. J.  sin.j?         j 

If  the  first  equation  in  ( W)  be  divided  by  the  first  in 
( V ),  we  shall  have, 

tan  ia  =  /  -cos.ff  cos.(S-A)       Y  J 
*2        \cos.(^~  B)cos.(S—  C)) 

And  corresponding  expressions  may  be  obtained  for 
tan.JS  and  tan.Jc. 


NAPIER'S   ANALOGIES. 

If  the  value  of  cos.c,  expressed  in  the  third  equation 
of  group  (#),  Prop.  7,  be  substituted  for  cos.c,  in  the 
second  member  of  the  first  equation  of  the  same  group, 
we  have, 

cos.a  =  cos.a  cos.2b  -f  sin. a  sin. b  cos.6  cos.(7-f-  sin.6  sin.c  cos. J.; 

which,  by  writing  for  cos.?5  its  equal,  1  —  sin.2£,  becomes, 

cos.a=cos.a — cos.asin.26-J-sin.cz  sin. 6  cos. 6  cos.  C+sin.b  sin.c  cos.-4. 

Or,  0  =  — cos.a  sin.25-|-sin.a  sin.6  cos. b  cos.  C+  sin. b  sin.c  cos.  J.. 

Dividing  through  by  sin.  b,  and  transposing,  we  find, 

cos.JL  sin.c  =  cos.a  sin.6 — sin.a  cos.5  cos.  C; 

,                    A      cos.a  sin.6  —  sin.a  cos.5  cos.  (7     , ... 
hence,  cos. A  = : — . .     ( 1 ) 

sin.c 

By  substituting  the  value  of  cos.c,  in  the  second  of  the 
equations  of  group  (S),  Prop.  7;  or,  more  simply,  by 
writing  B  for  J.,  and  b  for  a,  in  the  above  value,  for 
cos.^.,  we  obtain, 

-r>     eos.b  sin.a — sin.5  cos.a  cos. (7      ,ON 

COS.i*  = ; .         (2) 

sin.c 


SECTION    III.  349 

Adding  equations  (1)  and  (2),  member  to  member, 

we  have, 

.   .          ^     sin.fa+6)  —  sin. (a 4- b)  cos. 0 
C0S..A  +  C0S.2?  = a 1 . 1 L- ;    , 

sin,(? 

by  remembering  that  sin.a  cos.5  +  cos.a  sin.5  =  sin.(a+5). 
(See  Eq.  (7  ),  Sec.  I,  Plane  Trig.). 

Whence,  cos.J.  +  cos.£  =  (1 — cos.  C)  E^L±3m     ( 3 ) 

In  any  spherical  triangle  we  have,  (Prop.  I), 

sin.A  :  sin.i?  : :  sin.a  :  sin.5 ; 

And  therefore,  sin.J.  +  sin.i?  :  sin.jS  ::  sin.a  +  sin.6  : 

sin.5. 

-,-r  -      a   ,    -     t>     (sin.a  +  sin. b)  sin.i? 

Hence,  sm. A  +  smJ  =  i : — ~-l . 

sin.  6 

t,   ,    sin.i?      sin.  0     i  .  -.       i         n  sin.i?  .     ,,       , 

But,  — =  — — ,  which  value  of  — ; ,  in  the  above 

sm.6       sm.c  sm.6 

equation,  gives 

A   .    .     r>     (sin.a-f  sin.5)  sin. (7         , A, 

sm.  A  +  sinJ  =  i ■ '- .         ( 4 ) 

sm.c 

Dividing  equation  (4)  by  equation  (3)?  member  by 
member,  we  obtain, 

sin.  J.  +  sin.i?         sin.  0         sin.a  +  sin.  b        ,  -  v 

== x .       ( 5 ) 

cos.  A  +  cos.i?      1 — cos.  0       sin.(a  +  5) 

Comparing  this  equation  with  Equations  (20)  and  (26), 
Sec.  I,  Plane  Trigonometry,  we  see  that  it  can  be  re- 
duced to 

,       i  /  a   i   t>\         j.  i  n     sin.a-f  sin.5         ,nx 
tan.}(J.  +  j£)  =  cot.i(7x  -.    /"        -         (6) 

sin.(a  -f  b) 

Again,  from  the  proportion, 

sin. A  :  sin.i?  : :  sin.a  :  sin.5, 
we  likewise  have, 

sin. A — sin.J5  :  sin.I?  ::  sin.a — sin.5  :  sin.5; 

30    . 


350  SPHERICAL    TRIGONOMETRY. 

hence,    sin. J.  —  sin.i?  =  (sin.a  —  sin. b)  -.-—   =  (sin.a  — 

sin.  b 

.     ,N  sin.  0 

sin.  b)  — . 

sm.<? 

Dividing  this  equation  hy  equation  (3),  member  by 

member,  we  obtain, 

sin.J.  —  sin.i?  _      sin.C         sin.a — sin. b 

qoq.A  +  gos.B      1 — cos.  0       s'm.(a  +  b) 

Comparing  this  with  Equations  (22)  and  (26),  Sec.  I, 

Plane  Trigonometry,  we  see  that  it  will  reduce  to 

t™.i(A-£)  =  cotiCx    sin.(a  +  6).      (?) 

N"ow,sin.a  +  sin.5  =  2sin.(-^—  )cos.(— ^— );  Eq.  (15), 
Sec.  I,  Plane  Trig.). 

and,  sin.  (a  +  b)  =  2sin.(-+-)  cos.(^-5) ;  Eq.  (30), 

Sec.  I,  Plane  Trig.). 

Dividing  the  first  of  these  bj  the  second,  we  have 

,a —  b\ 

.     •     7       cos.( — s— ] 

sm.a  H-  sin. b  -  .  V     2    / 

sin.(a  +  b)      '         7a  +  b\ 

"Writing  the  second  member  of  this  equation  for  its 
first  member  in  Eq   (6),  that  equation  becomes 

tan.  l(A  +  B)  =  cot.  |g§glMj!^g.     ( 8 ) 

V  COS.  J(tf+&) 

By  a  similar  operation,  Eq.  (7)  may  be  reduced  to 

tan.  UA  -  JK  =  cot.  j(78m^(a~"^.     ( 9 ) 
2V  ;  2     sin.J(a+6) 

Equations  (8)  and  (9)  maybe  resolved  into  the  pro- 
portions 
cos.  J(a  +  b)  :  cos.  J(#  —  b)  : :  cot.  \Q  :  tan.  \(A  -f  i?) ; 
sin.  \(a  -f  5)  :  sin.  \{a  —  b)  : :  cot.  J(7  :  tan.  J(J. —  B). 
These  proportions  are  known  as  Napier's  1st  and  2d 


SECTION    III.  351 

Analogies,  and  may  be  advantageously  used  in  the  solu- 
tion of  spherical  triangles,  when  two  sides  and  the  in- 
cluded angle  are  given. 

When  expressed  in  language,  these  proportions  fur- 
nish the  following  rules : 

1.  The  cosine  of  the  half  sum  of  any  two  sides  of  a  spheri- 
cal triangle  is  to  the  cosine  of  the  half  difference  of  the  same 
sides,  as  the  cotangent  of  half  the  included  angle  is  to  the 
tangent  of  the  half  sum  of  the  other  two  angles. 

2.  The  sine  of  the  half  sum  of  any  two  sides  of  a  spheri- 
cal triangle  is  to  the  sine  of  the  half  difference  of  the  same 
sides,  as  the  cotangent  of  half  the  included  angle  is  to  the 
tangent  of  the  half  difference  of  the  other  two  angles. 

The  half  sum,  and  the  half  difference  of  two  angles 
of  a  spherical  triangle,  may  be  found  by  these  rules,  when 
two  sides  and  the  included  angle  are  given ;  and  by  add- 
ing the  half  sum  to  the  half  difference,  we  get  the 
greater  of  these  two  angles,  and  by  subtracting  the  half 
difference  from  the  half  sum,  we  get  the  smaller.  The 
third  side  may  then  be  found  by  proportion. 

We  have  analogous  proportions  applicable  to  the  case 
in  which  two  angles  and  the  included  side  of  a  spherical 
triangle  are  given. 

;  To  deduce  these,  let  us  represent  the  angles  of  the  tri- 
angle by  A,  B,  and  C,  and  the  opposite  sides  by  a,  b,  and 
c ;  A',  Bf,  (7,  a',  bf,  c',  denoting  the  corresponding  angles 
and  sides  of  the  polar  triangle. 

Now,  Eq.  ( 9 )  is  applicable  to  any  spherical  triangle, 
and  when  applied  to  the  polar  triangle,  it  becomes 

tan.  UA'  -B')  -  cot.  0  l^l^S^.  (») 

*v  '  2      sin.  \{a!  +  V) 

But  by  Prop.  6,  Sec.  I,   Spherical  Geometry,  we  have 

Af  =  180°  —  a,  .£'  =  180°  —  b,C'  =  180°  — .  c, 

a'  =  180°  —  A,  V  =  180°  —  B,  c'  =  180°  —  C. 

Whence,  l(Af-B>)=l(b-a\  \(a>:  +  V)  =  180°-  ^±^9 
j(a'  -  b')  =  i(B  -  A\  \C  =  90°  -  \c. 


352  SPHERICAL    TRIGONOMETRY. 

By  the  substitution  of  these  values  in  Eq.  (™),  that 
equation  becomes 

or,        tan.  J(a  _  b)  =  gj^fj  tan.  \c,         (p) 

since  tan.  J(5  —  a)  =  —  tan.  J(a  —  b\  and  sin.  J(i?  —  A  = 
—  sin.|(J.—  £). 

By  applying  Eq.  (8)  to  the  polar  triangle,  and  treating 
the  resulting  equation  in  a  manner  similar  to  the  above, 
we  find 

Equations  {p)  and  (q)  maybe  resolved  into  the  fol- 
lowing proportions. 

sin.  l(A  +  B)  :  sin.  %{A  —  B)  : :  tan.  \c  :  tan.  J(a  —  6); 
cos.  }(JL  +  J5)  :  cos.  J(J.  —  i?)  : :  tan.  \c  :  tan.  J(a  -f  b). 

These  proportions  are  called  Napier's  3d  and  4th 
Analogies,  and  when  expressed  in  words  become  the  fol- 
lowing rules : 

1.  The  cosine  of  the  half  sum  of  any  two  angles  of  a 
spherical  triangle  is  to  the  cosine  of  the  half  difference  of  the 
same  angles,  as  the  tangent  of  half  the  included  side  is  to  the 
tangent  of  the  half  sum  of  the  other  two  sides, 

2.  The  sine  of  the  half  sum  of  any  two  angles  of  a  spheri- 
cal triangle  is  to  the  sine  of  the  half  difference  of  the  same 
angles,  as  the  tangent  of  half  the  included  side  is  to  the  tan- 
gent of  the  half  difference  of  the  other  two  sides. 

The  half  sum,  and  the  half  difference  of  two  sides  of 
a  spherical  triangle,  may  be  found  by  these  rules,  when 
two  angles  and  the  included  side  are  given ;  and  by  add- 
ing the  half  sum  to  the  half  difference,  we  get  the  greater 
of  these  sides,  and  by  subtracting  the  half  difference 
from  the  half  sum,  we  get  the  smaller. 


* 


SECTION   IV.  353 


SECTION  IV. 


SPHERICAL  TRIGONOMETRY  APPLIED. 
SOLUTION  OF  RIGHT-ANGLED  SPHERICAL  TRIANGLES. 

A  good  general  conception  of  the  sphere  is  essential 
to  a  practical  knowledge  of  spherical  trigonometry,  and 
this  conception  is  best  obtained  by  the  examination  of 
an  artificial  globe.  By  tracing  out  upon  its  surface  the 
various  forms  of  right-angled  and  oblique-angled  tri- 
angles, and  viewing  them  from  different  points,  we  may 
soon  acquire  the  power  of  making  a  natural  representa- 
tion of  them  on  paper,  which  will  be  found  of  much  as- 
sistance in  the  solution  and  interpretation  of  problems. 

For  instance,  suppose  one  side  of  a  right-angled 
spherical  triangle  to  be  56°,  and  the  angle  between  this 
side  and  the  hypotenuse  to  be  24°.  What  is  the  hypote- 
nuse, and  what  the  other  side  and  angle  ? 

A  person  might  solve  this  problem  by  the  application 
of  the  proper  equations  or  proportions,  without  really 
comprehending  it ;  that  is,  without  being  able  to  form  a 
distinct  notion  of  the  shape  of  the  triangle,  and  of  its 
relation  to  the  surface  of  the  sphere  on  which  it  is 
situated. 

If  we  refer  this  triangle  to  the  common  geographical 
globe,  the  side  56°  may  be  laid  off  on  the  equator,  or  on 
a  meridian.  In  the  first  case,  the  hypotenuse  will  be  the 
arc  of  a  great  circle  drawn  through  one  extremity  of  the 
side  56°,  above  or  below  the  equator,  and  making  with 
30*  x 


354 


SPHERICAL    TRIGONOMETRY. 


it  an  angle  of  24° ;  the  other  side  will  be  an  arc  of  a 
meridian.  In  the  second  case,  the  side  56°  falling  on  a 
meridian,  the  hypotenuse  will  be  the  arc  of  a  great  circle 
drawn  through  one  extremity  of  this  side,  on  the  right 
or  left  of  the  meridian,  and  making  with  it  an  angle  of 
24° ;  the  other  side  will  be  the  arc  of  a  great  circle,  at 
right  angles  to  the  meridian  in  which  the  given  side  lies. 

Generally  speaking,  the  apparent  form  of  a  spherical 
triangle,  and  consequently  the  manner  of  representing 
it  on  paper,  will  differ  with  the  position  assumed  for  the 
eye  in  viewing  it.  From  whatever  point  we  look  at  a 
sphere,  its  outline  is  a  perfect  circle  in  the  axis  of  which 
the  eye  is  situated;  and  when  the  eye  is,  as  will  be  here- 
after supposed,  at  an  infinite  distance,  this  circle  will  be 
a  great  circle  of  the  sphere.  All  great  circles  of  the 
sphere  whose  planes  pass  through  the  eye,  will  seem  to 
be  diameters  of  the  circle  which  represents  the  outline 
of  the  sphere. 

"We  will  now  suppose  the  eye  to  be  in  the  plane  of  the 
equator,  and  proceed  to  construct  our  triangle  on  paper. 

Let  the  great  circle, 
PAS  A',  represent  the  out- 
line of  the  sphere,  the  di- 
ameter AA'  the  equator, 
and  the  diameter  PS  the 
central  meridian,  or  the 
meridian  in  whose  plane 
the  eye  is  situated.  Let 
AB  =  56°,  represent  the 
given  side,  and  A  (7,making 
with  AB  the  angle  B A  (7= 
24°,  the  hypotenuse,  then  will  BO,  the  arc  of  a  meridian, 
be  the  other  side  at  right  angles  to  AB,  and  the  triangle, 
ABO,  corresponds  in  all  respects  to  the  given  triangle. 

Again,  measure  off  5Q°  from  P  to  Q,  draw  the  radius 
DQ,  make  the  arc  A'G-  equal  to  24°,  and  draw  the  quad- 
rant PRGr.  The  triangle  PQR  will  also  represent  the 
given  triangle  in  every  particular. 


SECTION   IV.  355 

We  know  from  the  construction,  that  D  V,  =  24°,  is 
greater  than  BO,  and  that  A 0  is  greater  than  AB,  that 
is,  greater  than  56°. 

In  like  manner,  we  know  that  A',  =  24°,  is  greater 
than  QB,  and  that  PB  is  greater  than  PQ,  because  PB 
is  more  nearly  equal  toPG,  =90°,  thanP#  is  to  PA,  =90°. 

For  illustration  and  explanation,  we  also  give  the  fol- 
lowing example : 

In  a  right-angled  spherical  triangle,  there  are  given, 
the  hypotenuse  equal  to  150°  33'  20",  the  angle  at  the 
base,  23°  27'  29",  to  find  the  base  and  the  perpendicular. 
Let  A1  BO  in  the  last  figure,  represent  the  triangle  in 
which  A'0  =  150°  33'  20",  the  L  BAfO=  23°  27'  29", 
and  the  sides  A'B  and  BO  are  required. 

This  problem  presents  a  right-angled  spherical  tri- 
angle, whose  base  and  hypotenuse  are  each  greater  than 
90° ;  and  in  cases  of  this  kind,  let  the  pupil  observe, 
that  the  b^ase  is  greater  than  the  hypotenuse,  and  the  oblique 
angle  opposite  the  base,  is  greater  than  a  right  angle.  In 
all  cases,  a  spherical  triangle  andits  supplemental  triangle 
make  a  lune.  It  is  180°  from  one  pole  to  its  opposite, 
whatever  great  circle  be  traversed.  It  is  180°  along  the 
equator  ABA',  and  also  180°  along  the  ecliptic  AOA'. 
The  lune  always  gives  two  triangles;  and  when  the 
sides  of  one  of  them  are  greater  than  90°,  we  take  the 
triangle  having  supplemental  sides ;  hence  in  this  case 
we  operate  on  the  triangle  ABO. 

A  O  is  greater  than  AB,  therefore  A'B  is  greater  than 
the  hypotenuse  A'C. 

The  [_AOB  is  less  than  90°;  therefore,  the  adjacent 
angle  A' OB  is  greater  than  90°,  the  two  together  being 
equal  to  two  right  angles. 

These  facts  are  technically  expressed,  by  saying,  that 
the  sides  and  opposite  angles  are  of  the  same  affection.* 

*  Same  affection :  that  is,  both  greater  or  both  less  than  90°.  Dif- 
ferent affection :  the  one  greater,  the  other  less  than  90°. 


356  SPHERICAL    TRIGONOMETRY. 

!Now,  if  the  two  sides  of  a  right-angled  spherical  triangle 
are  of  the  same  affection,  the  hypotenuse  will  be  less  than 
90° ;  and  if  of  different  affection,  the  hypotenuse  will  be 
greater  than  90°. 

If,  in  every  instance,  we  make  a  riatural  construction 
of  the  figure,  and  use  common  judgment,  it  will  be  im- 
possible to  doubt  whether  an  arc  must  be  taken  greater 
or  less  than  90°. 

We  will  now  solve  the  triangle  A  OB.  AO  =  180°  — 
150°  33'  20"  =  29°  26'  40". 

To  find  BO,  we  use  Eq.  (3)  or  (13),  Prop.  3,  Sec.  II., 

thus: 

b,  sin.  29°  26'  40"     .       9.691594 

A,  sin.  23°_2T_29^     .-      9.599984 
a,sin.  11°  17'     7"     .       9.291578 

To  find  AB,  we  use  equation  ( 1 )  or  ( 11 ),  thus : 

a,  tan.  11°  17'     7"     .       9.300016 
A,  cot.  23°  27'  29"     .     10.362674 

c,sin.  27°  22'  32"     .       9.662690 

180 • 

A'B=lte°  37'  28" 

PRACTICAL  PROBLEMS  IN  RIGHT-ANGLED    SPHERICAL 
TRIGONOMETRY. 

1.  In  the   right-angled  spherical 
triangle  ABO,  given  AB  =  118°  21'     a- 
4",  and  the  angle  A  =  23°  40'  12",  to 
find  the  other  parts.  "~B 

A       (AG,  116°  17'  45";  the  angle  O,  100°  59'  26"; 
I      and  BO,  21°  5'  42". 

2.  In  the  right-angled  spherical  triangle  ABO,  given 
AB  53°  14'  20",  and  the  angle  A  91°  25'  53",  to  find 
the  other  parts. 

A       (AO,  91°  4'  9";  the  angle  O,  53°  15'  8"; 
^n*'\      and  BO,  91°  47'  11". 


SECTION  IV.  357 

3.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  102°  50'  25",  and  the  angle  A  113°  14'  87",  to 
find  the  other  parts. 

A       (AC,  84°  51'  36";  the  angle  C,  101°  46"  57"; 
\      and  BC,  113°  46'  27". 

4.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  48°  24'  16",  and  BC  59°  38'  27",  to  find  the 
other  parts. 

A       (AC,  70°  23'  42";  the  angled,  66°  20'  40"; 
s  I      and  the  angle  (7,  52°  32'  55". 

5.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  151°  23'  9",  and  ^(7  16°  35'  14",  to  find  the 
other  parts. 

An§  (AC,  147°  16'  51";  the  angle  C,  117°  37'  21"; 
*  I      and  the  angle  A,  31°  52'  50". 

6.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  73°  4'  31",  and  AC  86°  12'  15,"  to  find  the  other 
parts. 

Am  (BC,  76°  51'  20";  the  angled,  77°  24'  23"; 
m'  \      and  the  angle  C,  73°  29'  40". 
'  7. 'In  the  right-angled  spherical  triangle  ABC,  given 
AC  118°   32'   12",   and  AB  47°   26'   35",  to  find  the 
other  parts. 

A       (BC,  134°  56f  20";  the  angle  A,  126°  19'  2"; 
m'  \      and  the  angle  C,  56°  58'  44". 

8.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  40°  18'  23",  and  AC  100'°  3'  7",  to  find  the 
other  parts. 

A      (  The  angle  A,  98°  38'  53" ;   the   angle 
^'  t      C,  40°  4'  6" ;  and  BC,  103°  13'  52". 

9.  In  the  right-angled  spherical  triangle  ABC,  given 
AC  61°  3'  22",  and  the  angle  A  49°  28'  12",  to  find 
the  other  parts. 

Ang  (  AB,  49°  36'  6" ;  the  angle  C,  60°  29'  19" ; 
U  '  I     and  BC,  41°  41'  32". 

10.  In  the  right-angled  spherical  triangle  ABC,  given 


358  SPHERICAL    TRIGONOMETRY. 

AB  29°  12'  50",  and  the  angle  0  37°  26'  21",  to  find 
the  other  parts. 

r  Ambiguous ;  the  angle  A,  65°  27'  58",  or  its 

Ans. <      supplement;  J. (7,  53°    24/  13",  or  its  sup- 

(     plement;  BO,  46°  55'  2",  or  its  supplement. 

11.  In  the  right-angled  spherical  triangle  ABC,  given 
AB  100°  10'  3",  and  the  angle  0  90°  14'  20",  to  find 
the  other  parts. 

cAO,  100°  9'  5bff,  or  its  supplement;  BO, 
Ans.-l      1°  19'  53",  or  its  supplement;  and  the 
L     angle  A,  1°  21'  8",  or  its  supplement. 

12.  In  the  right-angled  spherical  triangle  ABO,  given 
AB  54°  21'  35",  and  the  angle  (7  61°  2'  15",  to  find 
the  other  parts. 

cBO,  129°  28'  28",  or  its  supplement;   AO, 

Ans.l      111°  44'  34",  or  its  supplement;  and  the 

I     angle  A,  123°  47'  44",  or  its  supplement. 

13.  In  the  right-angled  spherical  triangle  ABO,  given 
AB  121°  26'  25",  and  the  angle  O  1110  14'  37",  to 
find  the  other  parts. 

rThe  angle  A,  136°  0'  3",  or  its  supplement; 
Ans.  i      AO,  66°  15'  38",  or  its  supplement;  and 
L     BO,  140°  30'  56",  or  its  supplement. 


QUADRANTAL    TRIANGLES 

The  solution  of  right-angled  spheri- 
cal triangles  includes,  also,  the  solu- 
tion of  quadr anted  triangles,  as  may  be 
seen  by  inspecting  the  adjoining  fig- 
ure. When  we  have  one  quadr antal 
triangle,  we  have  four,  which  with  one 
right-angled  triangle,  fill  up  the  whole  hemisphere. 

To  effect  the  solution  of  either  of  the  four  quadrantal 
triangles,  APO,  AP'O,  A!PO,  oxA'P'O,  it  is  sufficient 
to  solve  the  small  right-angled  spherical  triangle  ABO. 


SECTION    IV.  359 

To  the  half  lune  AP'B,  we  add  the  triangle  ABO, 
and  we  have  the  quadrantal  triangle  AP'Q',  and  hy  sub- 
tracting the  same  from  the  equal  half  lune  APB,  we 
have  the  quadrantal  triangle  PAC. 

When  we  have  the  side,  AC,  of  the  same  triangle,  we 
have  its  supplement,  A'C,  which  is  a  side  of  the  triangles 
A'PC,  and  A'P'C.  "When  we  have  the  side,  CB,  of 
the  small  triangle,  by  adding  it  to  9.0°,  we  have  P'C,  a 
side  of  the  triangle  A'P'C;  and  subtracting  it  from  90°, 
we  have  PC,  &  side  of  the  triangles  APC,  and  AP'C. 

PROBLEM    I. 

In  a  quadrantal  triangle,  there  are  given  the  quadrantal 
side,  90°,  a  side  adjacent,  42°  21',  and  the  angle  opposite 
this  last  side,  equal  to  36°  31'.     Required  the  other  parts. 

By  this  enumeration  we  cannot  decide  whether  the  triangle  APC 
or  AP'C,  is  the  one  required,  for  AC  —  42°  21'  belongs  equally 
to  both  triangles.     The  angle  APC  =  AP'C  =  36°  31'  =  AB. 

We  operate  wholly  on  the  triangle  ABC. 

To  find  the  angle  A,  call  it  the  middle  part. 
Then,  R  cos.  CAB  =  R  sin.iM  C  =  cot.  J.  C  tan.^5. 


cot.  AC     =     42°  21' 
tzn.AB     =     36°  31' 

10.040231 
9.869473 

cos.  CAS  =     35°  40' 51" 
90° 

9.909704 

PAC  =     54°  19'    9" 
PrAC  =  125° 40'  51" 

To  find  the  angle  C,  call  it  the  middle  part. 

R cos.  ACB  =  sin.  CAB  cos.  AB. 

sin.  CAB  =     35°  40'  51"  9.765869 

zoz.AB     =     36°  31'  .        9.905085 


cos. ACB  =     62°    2' 45"  9.670954 

180° 


ACP  «  A'CP'  «=  117°  57'  15" 


360  SPHERICAL    TRIGONOMETRY. 

To  find  the  side  BO,  call  it  the  middle  part. 
Rs'm.BC  =  tan.XB  cot.ACB. 


tsm.AB     =     36°  31'    0" 
cotACB  =     62°    2' 45" 

9.869473 
9.724835 

sin.BG     =     23°    8' 11" 
90° 

9.594308 

PC  =     66°  51'  49" 
PC  =  113°    8' 11" 

"We  now  have  all  the  sides,  and  all  the  angles  of  the 
four  triangles  in  question. 

PROBLEM  II. 

In  a  quadr anted  spherical  triangle,  having  given  the  quad- 
rantal  side,  90°,  an  adjacent  side,  115°  09',  and  the  included 
angle,  115°  55',  to  find  the  other  parts. 

This  enunciation  clearly  points  out 
the  particular  triangle  A'P'O.  A'P' 
=  90°  ;  and  conceive  A'C=  115°  09'. 
Then  the  angle  P'AfO  =  115°  bbf  = 
P'D. 

From  the  angle  PM/(7take  90°,  or 
PfA'B,  and  the  remainder  is  the  angle 
OAfD  =  BAO  =  25°  55'. 

We  here  again  operate  on  the  triangle  ABO.  A'O, 
taken  from  180°,  gives 

64°  51'  =  AO. 

To  find  BO,  we  call  it  the  middle  part. 

R  sin. 5(7=  sin. A  C  sin. B A  C. 

sin. AC     =    64°  51'  .        9.956744 

sin.BAC  =     25°  55'  .        9.640544 


sm.BC     =     23°  18'  19"     .        9.597288 
90° 


PC  =  113°  18'  19" 


SECTION    IV.  361 

To  find  AB,  we  call  it  the  middle  part. 

Rsm.AB  =  t&n.BC  cot.BAC. 

t^ixi.BC     =     23°  18'  19"     .         9.634251 

cot.BAC  =     25°  55'  .         9.313423 


sin.AB      =     62°  26'    8"     .        8.947674 
180° 


A'B  =  117°  33'  52"  =  the  angle  AIP'C. 

To  find  the  angle  0,  we  call  it  the  middle  part. 
E  cos.  G  =  cot.  J.  C  tan.^a 

cot. AQ    =     64°  51'  .        9.671634 

tan.^BC    =     23°  18'  19"      .        9.634251 


cos.  C       =     78°  .        9.305885 

180°  19'  53"      . 


FCA!  =  101°  40'    7" 

Thus  we  have  found  the  side  FC  =  113°  18'  19" -\ 

The  angle  A'F  O  =  117°  33'  52"  I  An*. 
"     FCA'  =  101°  40'    7") 

PRACTICAL   PROBLEMS. 

1.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 
90°,  a  side  adjacent,  67°  3',  and  the  included  angle,  49° 
18',  to  find  the  other  parts. 

c  The  remaining  side  is  53°  &  46" ;  the  angle 
Ans.   <      opposite  the  quadrantal  side,  108°  32'  27" ; 
I     and  the  remaining  angle,  60°  48'  54". 

2.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 
90°,  one  angle  adjacent,  118°  40'  36",  and  the  side  op- 
posite this  last-mentioned  angle,  113°  2;  28",  to  find  the 
other  parts. 

c  The  remaining  side  is  54°  38'  57" ;  the  angle 
Ans.   <      opposite,  51°  2'  35";  and  the  angle  opposite 
I     the  quadrantal  side  72°  26'  21". 

3.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 

31 


362  SPHERICAL    TRIGONOMETRY. 

90°,  and  the  two  adjacent  angles,  one  69°  13'  46",  the 
other  72°  12'  4",  to  find  the  other  parts. 

r  One  of  the  remaining  sides  is  70°  8'  39",  the 
Arts.   <      other  is  73°  17'  29",  and  the  angle  opposite 

I     the  quadrantal  side  is  96°  13'  23". 

4.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 
90°,  one  adjacent  side,  86°  14'  40",  and  the  angle  oppo- 
site to  that  side,  37°  12'  20",  to  find  the  other  parts. 

(  The  remaining  side  is  4°  43'  2" ;  the  angle  op- 
Ans.   <      posite,  2°  51'  23" ;  and  the  angle  opposite 
I     the  quadrantal  side,  142°  42'  2". 

5.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 
90°,  and  the  other  two  sides,  one  118°  32'  16",  the  other 
67°  48'  40",  to  find  the  other  parts  —  the  three  angles. 

rThe  angles  are  64°  32'  21",  121°  3'  40",  and 
Arts.   <      77°  11'  6" ;  the  greater  angle  opposite  the 
I     greater  side,  of  course. 

6.  In  a  quadrantal  triangle,  given  the  quadrantal  side, 
90°,  the  angle  opposite,  104°  41'  17",  and  one  adjacent 
side,  73°  21'  6",  to  find  the  other  parts. 

m       (  Eemaining  side,  49°  42'  18" ;  remaining 
U8'  1     angles,  47°  32'  39",  and  67°  5V  13". 


SOLUTION  OF  OBLIQUE-ANGLED  SPHERICAL  TRIANGLES. 

All  cases  of  oblique-angled  spherical  trigonometry 
may  be  solved  by  right-angled  Trigonometry,  except 
two ;  because  every  oblique-angled  spherical  triangle  is 
composed  of  the  sum,  or  the  difference,  of  two  right- 
angled  spherical  triangles. 

When  a  side  and  two  of  the  angles,  or  an  angle  and  two 
of  the  sides  are  given,  to  find  the  other  parts,  conform  to 
the  following  directions : 

Let  a  perpendicular  be  drawn  from  an  extremity  of  a 
given  side,  and  opposite  a  given  angle  or  its  supplement; 
this  will  form  two  right-angled  spherical  triangles ;  and 


SECTION  IT.  363 

one  of  them  will  have  its  hypotenuse  and  one  of  its  ad- 
jacent angles  given,  from  which  all  its  other  parts  can 
be  computed ;  and  some  of  these  parts  will  become  as 
known  parts  to  the  other  triangle,  from  which  all  its 
parts  can  be  computed. 

To  facilitate  these  computations,  we  here  give  a  sum- 
mary of  the  practical  truths  demonstrated  in  the  fore- 
going propositions. 

1.  The  sines  of  the  sides  of  spherical  triangles  are  propor- 
tional to  the  sines  of  their  opposite  angles. 

2.  The  sines  of  the  segments  of  the  base,  made  by  a  per- 
pendicular from  the  opposite  angle,  are  proportional  to  the 
cotangents  of  their  adjacent  angles. 

3.  The  cosines  of  the  segments  of  the  base  are  proportional 
to  the  cosines  of  the  adjacent  sides  of  the  triangle. 

4.  The  tangents  of  the  segments  of  the  base  are  reciprocally 
proportional  to  the  cotangents  of  the  segments  of  the  vertical 
angle. 

5.  The  cosines  of  the  angles  at  the  base  are  proportional 
to  the  sines  of  the  corresponding  segments  of  the  vertical 
angle. 

6.  The  cosines  of  the  segments  of  the  vertical  angle  are 
proportional  to  the  cotangents  of  the  adjoining  sides  of  the 
triangle. 

The  two  cases  in  which  right-angled  spherical  triangles 
are  not  used,  are, 

1st.  When  the  three  sides  are  given  to  find  the  angles ; 
and, 

2d.  "When  the  three  angles  are  given  to  find  the  sides. 

The  first  of  these  cases  is  the  most  important  of  all, 
and  for  that  reason  great  attention  has  been  given  to  it, 
and  two  series  of  equations,  (2*  and  Z7,  Prop.  7,  Sec.  Ill), 
have  been  deduced  to  facilitate  its  solution. 

As  heretofore,  let  AB  0  represent  any  triangle  whose 
angles  are  denoted  by  A,  B,  and  (7,  and  sides  by  a,  b, 


364  SPHERICAL    TRIGONOMETRY. 

and  c ;  the  side  a  being  opposite  |__  A,  the  side  b  oppo- 
site [__  B>  etc* 

EXAMPLES. 

1.  In  the  triangle  ^LS  (7,  a  =  70°  4' 18";  b  =  63°  21' 27"; 
and  e,  59°  16'  23" ;  required  the  angle  A. 

The  formula  for  this  is  the  first  equation  in  group  (T, 
Prop.  7,  Sec.  Ill),  which  is 


cos. 


A  _  ,R2  sm.Ssm.(S—a\% 
'  2        \         sin.6  sin.<?         / 

We  write  the  second  member  of  this  equation  thus 


■j 


(-^l)  (-^~)  (sin'^  ™-(#-«), 
\sin.6/  Vsm.c/  v  n 


showing  four  distinct  factors  under  the  radical. 

The  logarithm  corresponding  to  - — -,  is  that  of  sin. b 

r> 

subtracted  from  10;   and  of  - —  is  that  of  sin.<?  sub- 
sin.  <? 

tracted  from  10,  which  we  call  sin.complement. 

BC=a=    70°     4'  18" 

AB  =  c  =    59°  16'  23"  sin.  com.       .065697 

AC  =  b  =    63°  21'  27"  sin.  com.       .048749 


2)192°  42'  8" 
S   -  96°  21'  4"  sin. 
S  —  a  =  26°  16'  46"  sin. 

9.997326 
9.646158 

J  J.  —  40°  49'  10"  cos. 

2 

2  )  19.757930 

9.878965 

A  =  81°  38'  20" 

"When  we  apply  the  equation  to  find  the  angle  A,  we 
write  a  first,  at  the  top  of  the  column ;  when  we  apply 
the  equation  to  find  the  angle  B,  we  write  b  at  the  top 
of  the  column.     Thus, 


SECTION  IV.  365 

To  find  the  ande  B. 


*&■ 


cos.$b —sy  - 


sin. a  sin.c 


-V    IS-)  (^~)  (-sia-S^  sm.(S-b) 
V      Vsin.a/  Vsin.c/  v  y  y 

b  =  63°  21'  27" 

c  =  59°  16'  23"     sin.com.      .        .065697 

a=  70°     4'  18"     sin.com.     .         .026875 

2)192°  42'     8" 


S=  96°  21'     4"     sin.  .         .      9.997326 
£— .6  =  32°  59'  37"     sin..         .      9.736034 

2)19.825872 

}£*-  35°    4'  49"    cos.  .  9.912936 

2 


jg=  70°     9'  38" 

By  the  other  equation  in  formulae  (T,  Prop.  7,  Sec. 
Ill),  we  can  find  the  angle  O;  but,  for  the  sake  of  variety, 
we  will  find  the  angle  O  by  the  application  of  the  third 
equation  in  formulae  ( U9  Prop.  7,  Sec.  III). 


dn,iC-\  /^^•(^-&)  sin.(£-a) 
v  sin.6  sin.a 


c  =  59°  16'  23" 

am  70°     4'  18"     sin.com.         .026817 
b  =  63°  21'  27"     sin.com.         .048479 
2)192°  42'     8" 


£=96°  21'    4" 
S—  a  =  26°  16'  46"    sin.        .     9.646158 
S—  6  =  32°  59'  37"    sin.        .     9.736034 

2  )  19.4574~88 

J  a  =32°  23'  17"    sin.        .     9.778744 
2 


C=64°  46'  34" 
35*      . 


366  SPHERICAL    TRIGONOMETRY. 

To  show  the  harmony  and  practical  utility  of  these  two 
sets  of  equations,  we  will  find  the  angle  A,  from  the 
equation 

M = s/  {£i)  §§  siu-(^- 6)  sia-iS- c)- 

a  =  70°     4'    18" 

b  m  63°  21'    27"    sin.com.         .048749 

c  =  59°  16'    23"    sin.com.         .065697 


2  )  192°  42'  8" 

8m   96°  21'  4" 

JS—b=   32°  59'  37"  sin. 
#— c  =  370  4/  41"  sin# 

9.736034 

9.780247 

M  =  40°  49'  10"  sin. 

2 

2  )  19.630727 
9.815363 

^L  =  81°  38'  20" 
2.  In  a  spherical  triangle  ABC,  given  the  angle  J.,  38° 
19'  18";  the  angle  B,  48°  0'  10";    and  the  angle  O, 
121°  8'  6" ;  to  find  the  sides  a,  b,  c. 

By  passing  to  the  triangle  polar  to  this,  we  have, 
(Prop.  6,  Sec.  I,  Spherical  Geometry), 

A  =  38°  19'  18"  supplement  141°  40'  42" 
£=  48°  0'  10"  supplement  131°  59'  50" 
(7=  121°     8'     6"  supplement    58°  51'  54" 

We  now  find  the  angles  to  the   spherical  triangle, 
the  sides  of  which  are  these  supplements. 
Thus,       .       141°  40'  42" 


131°  59'  50" 

sin.com. 

.128909 

58°  51'  54" 

sin.com. 

.067551 

2  )  332°  32'  26" 

166°  16'  13" 

sin. 

9.375375 

24°  35'  31" 

sin. 

9.619253 
2)19.191088 

66°  47'  374" 

cos. 

9.595544 

SECTION  IV.  367 

60°  47'  37*" 

2^ 

angle  =  121°  35'  15" 

supp.  =    58°  24'  45"  =  a  of  the  original  triangle. 
In  the  same  manner  we  find  b  =  60°  14'  25";  c  =  89°  V  14". 

It  is  perhaps  better  to  avoid  this  indirect  process  of 
computing  the  sides  of  a  spherical  triangle  when  the 
angles  are  given,  by  the  application  of  the  equations  in 
group  V  or  W,  Prop.  8,  Sec.  III.  "We  will  illustrate 
their  use  by  applying  the  second  equation  in  group  ( W), 
for  computing  the  side  b.     This  equation  is 

—  cos.tf  cos.(£— B)\i 


sin.^6  =  (- 


sin. J.  sin.  C  J 

A  =  38°  19'  18" 
B  =  48°  0'  10" 
(7=121°    8'    6" 

2)207°  27'  34" 
£=103°  43' 47"— cos.£=  +  sin.l3°  43'  47"=  9.375376 
B  =    48°    O'lO"    cos.(£— B)  =  55°  43'  37"=  9.750612 
(£—.#)  =    55°  43' 37"  2)19.125988 

square  root  =  9.562994 
sin.J[=    38°  19'  18"  =9.792445 
sin.  C=  121°     8'    6"  =  9.932443 

2)19.724888 

square  root  =  9.862444  =  9.862444 

diff.  — 1.700550 
Add  10,  for  radius  of  the  table,     10 

Tabular  sm.\b  =    30°     7'  14"  =1) J00550 

2 


I  =  60°  14'  28",  nearly. 

PRACTICAL    PROBLEMS. 

1.  In  any  triangle,  ABC,  whose  sides  are  a,  5,  c,  given 
5  =  118°  2'  14",  em  120°  18'  33",  and  the  included  angle 
A  -  27°  22'  34",  to  find  the  other  parts. 


368 


SPHERICAL    TRIGONOMETRY. 


A       (  a  =  23°  57'  13",  angle  B  =  91°  26'  44,  and  0  = 
T^rl      102°  5' 54". 

2.  Given,  .4  -  81°  38'  17",  J5  =  70°  9'  38",  and  (7  = 
64°  46'  32",  to  find  the  sides  a,  b,  e. 

A       (  a  m  70°  4'  18",  b  =  63°  21'  27",  and  c  =  59°  16' 
AnS'\     23". 

3.  Given,  the  three  sides,  a  =  93°  27'  34",  b  =  100°  4' 
26",  and  e  =  96°  14'  50",  to  find  the  angles  A,  B,  and  O. 

A       (A  =  94°  39'  4",  £  =  100°  32'  19",  and  (7=  96° 
n&  t      58' 36". 

4.  Given,  two  sides,  6  =  84°  16',  c  =  81°  12',  and  the 
angle  C=  80°  28',  to  find  the  other  parts. 

rThe  result  is  ambiguous,  for  we  may  consider 
the  angle  B  as  acute  or  obtuse.  If  the  angle 
B  is  acute,  then  A  -  97°  13'  45'',  B  =  83°  11' 
24",  and  a  -  96°  13'  33".  If  B  is  obtuse,  then 
A  =  21°  16'  44",  B  =  96°  48'  36",  and  a  m 
21°  19'  29". 

5.  Given,  one  side,  c=64°'26',  and  the  angles  adjacent, 
A  =  49°,  and  B  =  52°,  to  find  the  other  parts. 

(b  =  45°  56'  46",  a  =  43°  29'  49",  and  (7=  98° 
*'l     28' 5". 

6.  Given,  the  three  sides,  a=  90°,  5=  90°,  c  =  90°,  to 
find  the  angles  A,  B,  and  O. 

Am.  A  =  90°,  B  =  90°,  and  (7=  90°. 

7.  Given,  the  two  sides,  a  =  77°  25'  11",  c  =  128°  13' 
47",  and  the  angle  C  =  131°  11'  12",  to  find  the  other 
parts. 

An^  (  b  =  84°  29'  24",  A  =  69°  14',  and  B  m  72°  28' 


Ans. 


n 


46". 


8.  Given,  the  three  sides,  a  =  68°  34'  13",  b  =  59° 
21'  18",  and  c  =  112°  16'  32",  to  find  the  angles  A,  B, 
and  C. 

Ans.  {  A  =  45°  26'  12">  B  =  41°  n/  6">  °  =  134°  54' 


SECTION    IV.  S69 

9.  Given,  a*=  89°  21'  87",  6==  97°  18'  39",  <?  =  86°  53' 
46",  to  find  A,  B,  and  0, 

An    (  A  =  88°  57'  20",  £  =  97°  21'  26",  (7  =  86°  47' 

10.  Given,  a  m  31°  26'  41",  c  =  43°  22'  13",  and  the 
angle  JL=12°  16',  to  find  the  other  parts. 

c Ambiguous;  b  =  73°  7'  35",  or  12°  17'  39"; 
Ana  J      angle  £=115°  0'  31",  or  47°  1'  36";  (7  =  16° 
I     14'  27",  or  163°  45'  33". 

11.  In  a  triangle,  ABO,  we  have  the  angle  A =56°  18' 
40",  .£  -  39°  10'  38";  AD,  one  of  the  segments  of  the 
base,  is  32°  54'  16".  The  point  D  falls  upon  the  base 
AB,  and  the  angle  C  is  obtuse.  Eequired  the  sides  of 
the  triangle  and  the  angle  (7. 

(  (7=135°  47'  56",  <?=123°  4'  56f/,  «=90°  8'  17", 
***\      b  =  49°  23'  41". 

12.  Given,  J.  =  80°  10'  10",  B  =  58°  48'  36",  (7  =  91° 
52'  42",  to  find  a,  b,  and  c. 

Arts,  a  =  79°  38'  21",  5  =  58°  39'  16",  <?  =  86°  12'  52". 


370 


SPHERICAL    TRIGONOMETRY. 


SECTION   V. 


APPLICATIONS  OF  SPHERICAL  TRIGONOMETRY  TO 
ASTRONOMY  AND   GEOGRAPHY. 

SPHERICAL  TRIGONOMETRY  APPLIED  TO  ASTRONOMY. 


Spherical  Trigonometry  becomes  a  science  of  incalcu- 
lable importance  in  its  connection  with  geography,  navi- 
gation, and  astronomy;  for  neither  of  these  subjects  can 
be  understood  without  it ;  and  to  stimulate  the  student 
to  a  study  of  the  science,  we  here  attempt  to  give  him  a 
glimpse  at  some  of  its  points  of  application. 

Let  the  lines  in  the 
annexed  figure  represent 
circles  in  the  heavens 
above  and  around  us. 

Let  Z  be  the  zenith,  or 
the  point  just  overhead, 
Hch  the  horizon,  PZR 
the  meridian  in  the  hea- 
vens, and  P  the  pole  of 
the  earth's  equator;  then 
Ph  is  the  latitude  of  the 
observer,  and  PZ  is  the 

co.latitude.  Qcq  is  a  portion  of  the  equator,  and  the 
dotted,  curved  line,  mSfS,  parallel  to  the  equator,  is  the 
parallel  of  the  sun's  declination  at  some  particular  time ; 
and  in  this  figure  the  sun's  declination  is  supposed  to  be 
north.    By  the  revolution  of  the  earth  on  its  axis,  the 


SECTION   V.  371 

sun  is  apparently  brought  from  the  horizon,  at  S,  to  the 
meridian,  at  m;  and  from  thence  it  is  carried  down  on 
the  same  curve,  on  the  other  side  of  the  meridian ;  and 
this  apparent  motion  of  the  sun  (or  of  any  other  celestial 
body,)  makes  angles  at  the  pole  P,  which  are  in  direct 
proportion  to  their  times  of  description. 

The  apparent  straight  line,  Zc,  is  what  is  denominated, 
in  astronomy,  the  prime  vertical;  that  is,  the  east  and  west 
line  through  the  zenith,  passing  through  the  east  and  west 
points  in  the  horizon. 

When  the  latitude  of  the  place  is  north,  and  the  decli- 
nation is  also  north,  as  is  represented  in  this  figure,  the 
sun  rises  and  sets  on  the  horizon  to  the  north  of  the  east 
and  west  points,  and  the  distance  is  measured  by  the  arc, 
cS,  on  the  horizon. 

This  arc  can  be  found  by  means  of  the  right-angled 
spherical  triangle  cqS,  right-angled  at  q.  Sq  is  the  sun's 
declination,  and  the  angle  Scq  is  equal  to  the  co. latitude 
of  the  place ;  for  the  angle  Pch  is  the  latitude,  and  the 
angle  Scq  is  its  complement. 

The  side  cq,  a  portion  of  the  equator,  measures  the 
angle  cPq,  the  time  of  the  sun's  rising  or  setting  before 
or  after  six  o'clock,  apparent  time.  Thus  we  perceive  that 
this  little  triangle,  cSq,  is  a  very  important  one. 

"When  the  sun  is  exactly  east  or  west,  it  can  be  deter- 
mined by  the  triangle  ZPS' ';  the  side  PZ  is  known, 
being  the  co.latitude ;  the  angle  PZSf  is  a  right  angle, 
and  the  side  PSr  is  the  sun's  polar  distance.  Here,  then, 
are  the  hypotenuse  and  side  of  a  right-angled  spherical 
triangle  given,  from  which  the  other  parts  can  be  com- 
puted. The  angle  ZPS'  is  the  time  from  noon,  and  the 
side  ZSf  is  the  sun's  zenith  distance  at  that  time. 

The  following  problems  are  given,  to  illustrate  the 
important  applications  that  can  be  made  of  the  right- 
angled  triangle  cqS. 


372  SPHERICAL    TRIGONOMETRY. 

PRACTICAL    PROBLEMS. 

1.  At  what  time  will  the  sun  rise  and  set  in  Lat.  48° 
"N,,  when  its  declination  is  21°  N".  ? 

In  this  problem,  we  must  make  qS =21°,  PA  =  48°=the  angle 
Pch.  Then  the  angle  Scq  =  42°.  It  is  required  to  find  the  arc 
cq,  and  convert  it  into  time  at  the  rate  of  four  minutes  to  a  degree. 
This  will  give  the  apparent  time  after  six  o'clock  that  the  sun  sets, 
and  the  apparent  time  before  six  o'clock  that  the  sun  rises,  (no 
allowance  being  made  for  refraction). 
Making  cq  the  middle  part,  we  have 

R  sin.cq  =  tan.21°  tan.48° 
tan.21°  =  9.584177 
tan.48°  =  10.045563 


sin.c2=25°14'  5"  = 

=  25.2346° 
4 

Adding  to 

1A  40m  56' 
6* 

Sun  sets  p.  M., 

7*  40™  56*, 

From 
Taking 

6* 

1A  40m  56' 

Sun  rises  A.M., 

4*19™    4', 

9.629740,  rejecting  10. 


4',  apparent  time. 
From  this  we  derive  the  following  rule  for  finding  the  apparent 
time  of  sunrise  and  sunset,  assuming  that  the  declination  under- 
goes no  change  in  the  interval  between  these  instants,  which  we 
may  do  without  much  error. 

RULE. 

To  the  logarithmic  tangent  of  the  sun's  declination,  add  the 
logarithmic  tangent  of  the  latitude  of  the  observer  ;  and,  after 
rejecting  ten  from  the  result,  find  from  the  tables  the  arc  of 
which  this  is  the  logarithmic  sine,  and  convert  it  into  time  at 
the  rate  of  4  minutes  to  a  degree. 

This  time,  added  to  6  o'clock,  will  give  the  time  of  sunset, 
and,  subtracted  from  6  o'clock,  will  give  the  time  of  sunrise, 


SECTION   V.  373 

when  the  latitude  and  declination  are  both  north  or  both 
south  ;  but  when  one  is  north,  and  the  other  south,  the  addi- 
tion gives  the  time  of  sunrise,  and  the  subtraction  the  time  of 
sunset. 

2.  At  what  time  will  the  sun  set  when  its  declination 
is  23°  12'  K,  and  the  latitude  of  the  place  is  42°  40'  K  ? 

Ans.  lh  33M  8s,  apparent  time. 

3.  What  will  be  the  time  of  sunset  for  places  whose 
latitude  is  42°  40'  K.,  when  the  sun's  declination  is  15° 
21'  south  ?      •  Ans.  5h  lm  20s,  apparent  time. 

4.  What  will  be  the  time  of  sunrise  and  sunset  for 
places  whose  latitude  is  52°  30'  !N\,  when  the  sun's  decli- 
nation is  18°  42'  south  ? 

A         f  Rises  7h  44m  42s, )  ... 

An8'    (Sets    4ft  15m  18/ |  apparent  time. 

5.  What  will  be  the  time  of  sunset  and  of  sunrise  at 
St.  Petersburgh,  in  lat.  59°  56',  north,  when  the  sun's 
declination  is  23°  24',  north?  What  will  be  its  ampli- 
tude at  these  instants  ?  Also,  at  what  hours  will  it  be 
#ue  east  and  west,  and  what  will  be  its  altitude  at  such 
times  ? 

Sun  sets  at  9*  13m  30*  p.m.  \  apparent 
Sun  rises  at  2*  46m  30*  a.m.  J       time. 
Sun  rises  N.  of  east  1  ro0  9c/  on// 
Ans.  i  Sun  sets  ST.  of  west   J 

Sun  is  east  at  6*  58OT  a.m. 
Sun  is  west  at  5h  2m  p.m. 
I  Alt.  when  east  and  west  is  27°  19'. 


ON  THE  APPLICATION  OF  OBLIQUE-ANGLED   SPHERICAL 
TRIANGLES. 

One  of  the  most  important  problems  in  navigation 
and  astronomy,  is  the  determination  of  the  formula  for 
32 


374 


SPHEKICAL   TRIGONOMETRY. 


time.  This  problem  will 
be  understood  by  the  tri- 
angle PZS.  When  the 
sun  is  on  the  meridian,  it 
is  then  apparent  noon. 
When  not  on  the  meri- 
dian, we  can  determine 
the  interval  from  noon, 
by  means  of  the  triangle 
PZS;  for  we  can  know 
all  its  sides;  and  the 
angle  at  P,  changed  into 

time  at  the  rate  of  15°  to  one  hour,  will  give  the  time 
from  apparent  noon,  when  any  particular  altitude,  as 
TS,  may  have  been  observed.  PS  is  known,  by  the  sun's 
declination  at  about  the  time ;  and  PZ  is  known,  if  the 
observer  knows  his  latitude. 

Having  these  three  sides,  we  can  always  find  the  sought 
angle  at  the  pole,  by  the  equations  already  given  in 
formulae  (T,  or  U,  Prop.  7,  Sec.  Ill);  but  these  formulae 
require  the  use  of  the  co. latitude  and  the  co.altitude,  and 
the  practical  navigator  is  very  averse  to  taking  the  trou- 
ble of  finding  the  complements  of  arcs,  when  he  is  quite 
certain  that  formulae  can  be  made,  comprising  but  the 
arcs  themselves. 

The  practical  man,  also,  very  properly  demands  the 
most  concise  practical  results.  No  matter  how  much 
labor  is  spent  in  theorizing,  provided  we  arrive  at  prac- 
tical brevity ;  and  for  the  especial  accommodation  of 
seamen,  the  following  formula  for  finding  time  has  been 
deduced. 

From  the  symmetrical  formulae  (*'),  Prop.  7,  Sec.  HE, 
we  have, 

cos.ZS—  eos.PZ  cos.  PS 


cos.P  = 


sm.PZ  sm.PS 


Now,  in  place  of  cos.ZS,  we  take  siii.aST,  which  is,  in 


SECTION    V.  375 

fact,  the  same  thing ;  and  in  place  of  cos.PZ,  we  take 
sin.lat.,  which  is  also  the  same. 

In  short,  let  A  =  the  altitude  of  the  sun,  L  =  the  la- 
titude of  the  observer,  and  D  =  the  sun's  polar  distance. 

Then,      cos.P=!iBl4=s_in^4?^ 
C08.L  sm.D 

But,    2sin.2|P  =  1— -cos.P.      (See  Eq.  32,  Prop.  2, 
Sec.  I,  Plane  Trig.) 
Therefore, 

n.    51D      w       sin.J. — sin.L  cos.D 
2sm.2  JP  =  1  — —  

cos.L  sin.x> 

_  (cos.L  sin._D  +  sin.Z  cos.D)  —  sin.  A 

coa.L  sin.2) 

_  sin.(2y  4-  D)  —  sin. A 

cos.L  sin.i) 

Considering  (L  +  D)  as  a  single  arc,  and   (applying 

Equation  16,  Sec.  I,  Plane  Trig.),  we  have,  after  dividing 

bj  2, 

(L  +  D  +  A\    .    (L  +  2>  —  A\ 


cos, 


sin.JP  = 

cos.xe  sm.x> 

and  if  we  assume  S  = g , 

A 

,    ,,  ,  •    oir»      cos.aS'  sin.(/Sr — A) 

we  shall  have,  sin.*  IP  = =*A — H — J- 

cos.L  sm.D 


Or,  sin.JP 


.    /cos.S  sin.(ff — A) 
V        cos.X  sin.i) 


This  is  the  final  result,  when  the  radius  is  unity ;  and 
when  the  radius  is  greater  by  B,  then  the  sin.  JP  will  be 
greater  by  R ;  and,  therefore,  the  value  of  this  sine,  cor- 
responding to  our  tables,  is, 

sin.JP  =  \J(J*\  (-JL>j  cos.tfsin.^—  A). 
v    Vcos.iy/  Vsin.i)/ 


376  SPHERICAL    TRIGONOMETRY. 

PRACTICAL  PROBLEMS. 

1.  In  lat.  39°  6'  20"  North,  when  the  sun's  declination 
was  12°  3'  10"  North,  the  true  altitude*  of  the  sun's  cen- 
ter was  observed  to  be  30°  10'  40",  rising.  What  was 
the  apparent  time  ? 


ait    30°  10'  30" 
Lat.  39°  6' 20" 
RD.   77°  56'  50" 

2  )  147°  13'  40" 

S  =  73°  36'  50" 

cos.com.  .110146 
sin.com.  .009680 

cos.    9.450416 

tr- 

- A)   =  43°  26'  20" 

30°  22'  5" 

2 

sin.    9.837299 

2  )  19.407541 
.  sin.    9.703770 

P  =  60°  44'  10" 

This  angle,  converted  into  time  at  the  rate  of  15°  to 
one  hour,  or  4  minutes  to  1°,  gives  4*  2m  56'  from  appa- 
rent noon;  and  as  the  sun  was  rising,  it  was  before 
noon  or 

If  to  this  the  equation  of  time  were  applied,  we  should 
have  the  mean  time ;  and  if  such  time  were  compared 
with  that  of  a  clock  or  watch,  we  could  determine  its 
error.  A  good  observer,  with  a  good  instrument,  can, 
in  this  manner,  determine  the  local  time  within  4  or  5 
seconds. 

2.  In  lat.  40°  21'  North,  the  true  altitude  of  the  sun,  in 
the  forenoon,  was  found  to  be  36°  12',  when  the  declina- 

*  The  instrument  used,  the  manner  of  taking  the  altitude,  its  cor- 
rection for  refraction,  semi-diameter,  and  other  practical  or  circum- 
stantial details,  do  not  belong  to  a  work  of  this  kind,  but  to  a  work  on 
Practical  Astronomy  or  Navigation. 


SECTION  V.  377 

tion  of  the  sun  was  3°  20'  South.     "What  was  the  appa- 
rent time  ?  Ans.  9A  43m  44*  a.  m. 

3.  In  latitude  21°  2'  South,  when  the  sun's  declination 
was  18°  32'  North,  the  true  altitude,  in  the  afternoon, 
was  found  to  be  40°  8'.  What  was  the  apparent  time 
of  day?  Ans.  2A  2m  p.  m. 


SPHERICAL  TRIGONOMETRY  APPLIED  TO  GEOGRAPHY. 

If  we  wish  to  find  the  shortest  distance  between  two 
places  over  the  surface  of  the  earth,  when  the  dis- 
tance is  considerable,  we  must  employ  Spherical  Trigo- 
nometry. 

Suppose  the  least  distance  between  Rome  and  New 
Orleans  is  required ;  we  would  first  find  the  distance  in 
degrees  and  parts  of  a  degree,  and  then  multiply  that 
distance  by  the  number  of  miles  in  one  degree. 

In  the  solution  of  this  problem,  it  is  supposed  that  we 
have  the  latitude  and  longitude  of  both  places.  Then 
the  distances,  in  degrees,  from  the  north  pole  of  the 
earth  to  Rome  and  to  New  Orleans  are  the  two  sides  of 
a  spherical  triangle,  the  difference  of  longitude  of  the 
two  places  is  the  angle  at  the  pole  included  between 
these  sides,  and  the  problem  is,  to  determine  the  third 
side  of  a  spherical  triangle,  when  wre  have  two  sides  and 
the  included  angle  given. 

Let  P  be  the  north  pole,  B  the  position  of  Rome,  and 
N  that  of  New  Orleans. 

Lat.  Long. 

New  Orleans,  29°  57'  30"  N.  90°  "W. 

Rome,  41°  53'  54"  N.  12°  28'  40"  E. 

Whence,  PR  m  48°  6'    6", 

pjjf  =  60°  2'  30". 

Angle  NPR  =  102°  28'  40". 

32* 


378 


SPHERICAL  TRIGONOMETRY. 


We  now  employ  Na- 
pier's 1st  and  2d  Analo- 
gies, and  find  the  dis- 
tance, in  degrees,  to  be 
101°  31'  30".  This  re- 
duced to  miles,  at  the 
rate  of  69.16   miles  to     I  \ 

the  degree,  will  make     \  / 

the    distance   7021.469      \  j 

miles.  \  / 

The  angle  at  JV  is 
47°  49',  and  at  B,  59°              \^  ^"' 

35'  40".  ^ '"' 

The  third  side  of  a  spherical  triangle  can  be  found  by 
a  single  formula,  as  we  shall  see  by  inspecting  formulae 
(£')  Prop.  7,  Sec.  III. 

Let  0  be  the  included  angle,  and  c  the  unknown  side 
opposite ;  then, 

„        cos.c  —  cos.a  cos.5 

cos.  C  =  ^ r— i 

sin.a  sin. 6 

Adding  1  to  each  member,  and  reducing,  observing  at 
the  same  time  that  1  -j-  cos.  (7=  2cos.2  J  (7,  we  have, 
0       «...  ~     sin.a  sin.5  —  cos.a  cos.5  +  cos.e 

2COS.2  \  C  = : : 

siu.a  sm.6 
Whence,  2cos.2J(7  sin.a  sin.6  =  cos.c  —  cos.(a-f  b); 
or,  co8.c  =  cos. (a  -f  b)  +  2cos.2 \Q  sin.a  sin.6. 

The  second  member  of  this  equation  is  the  algebraic 
sum  of  two  decimal  fractions,  and  expresses  the  value  of 
the  natural  cosine  of  the  side  sought. 

This  case  of  Spherical  Trigonometry,  namely,  that  in 
which  two  sides  and  the  included  angle  are  given,  to 
find  the  third  side,  is  very  extensively  used  in  practical 
astronomy,  in  finding  the  angular  distance  of  the  moon 
from  the  sun,  stars,  and  planets.  For  this  purpose,  the 
right  ascension  and  declination  of  each  body  must  be 


SECTION    V. 


3T9 


found  for  the  same  moment  of  absolute  time.  Their 
difference  in  right  ascen- 
sion gives  the  included 
angle,  P,  at  the  celestial 
pole.  The  declination 
subtracted  from  90°,  if  it 
be  north,  and  added  to 
90°,  if  it  be  south,  will 
give  the  sides,  PZ  and 
PS. 

In  the  following  exam- 
ples, we  give  the  right 
ascension  and  declination 
of  the  bodies,  and  from 

these  the  student  is  required  to  compute  the  distance 
between  them. 

The  right  ascensions  are  given  in  time.     Their  differ- 
ence must  be  changed  to  degrees  for  the  included  angle. 


June  24,  1860. 

MEAN   TIME    GREENWICH. 


moon's 
R.  A.  Dec. 

h.   m.    e.  •:*      " 

At  noon,  10  51  36.5  3  33  24  N. 

«  3  h.,    10  58     1  2  47  43 

"  6  h.,    11     4  24.6  1  59  56.2 

"  9  h.,    11  10  47.6  1  12    6 


JUPITER'S 


R.  A. 
m.  s. 
4  27.6 
4  34.2 
4  40.8 
4  47.2 


Dec. 

o      t       it 

20  51  36.8  N. 
20  51  17.8 
20  50  58.7 
20  50  39.6 


Distance. 
o       t      tf 

44  8  12 

45  53  47 
47  39  18 
49  24  43 


October  6,  1860. 


)  R.A. 
h.  m.  s. 
At  noon,  5  41  21.8 
"  3  h.,  5  48  30.1 
"  6  h.,  5  55  40 
6  2  50.5 
6  10  1.2 


"  9  h., 
"  12  h., 


Dec. 

O  R.A. 

Oil! 

h.  m.  s. 

26  8  ON. 

12  49  27.4 

26  3  20 

12  49  54.8 

25  57  19 

12  50  22.2 

25  49  58 

12  50  49.6 

25  41  15.8 

12  51  11.9 

Dec. 

Q        I  II 

5  18  31  S. 

5  20  13.7 

5  21  56.4 

5  23  38.1 

5  25  20.8 


Distance. 

o         I         If 

107  37  2 
106  8  19 
104  39  19 
103  10  0 
101  40  23 


380  SPHERICAL   TRIGONOMETRY 


SECTION  VI 


REGULAR  POLYEDRONS. 

A  Regular  Polyedron  is  a  polyedron  having  all  its  faces  equal 
and  regular  polygons,  and  all  its  polyedral  angles  equal. 

The  sum  of  all  the  plane  angles  bounding  any  polyedral  angle  is 
less  than  four  right  angles ;  and  as  the  angle  of  the  equilateral  tri- 
angle is  |  of  a  right  angle,  we  have  f  x  3<4,  |  x  4<[4,  and  |  x  5<4 ; 
but  |  x  6=4,  |  x  7]>4,  and  so  on.  Hence,  it  follows  that  three, 
and  only  three,  polyedral  angles  may  be  formed,  having  the  equi- 
lateral triangle  for  faces;  namely,  a  triedral  angle  and  polyedral 
angles  of  four  and  of  five  faces. 

There  are,  therefore,  three  distinct  regular  polyedrons  bounded 
by  the  equilateral  triangle. 

1.  The  Tetraedron,  having  four  faces  and  four  solid  angles. 

2.  The  Octaedron,  having  eight  faces  and  six  solid  angles. 

3.  The  Icosaedron,  having  twenty  faces  and  twenty  solid  angles. 
With  right  plane  angles  we  can  form  only  a  triedral  angle ;  hence, 

with  equal  squares  we  may  bound  a  solid  having  six  faces  and  eight 
equal  triedral  angles.     This  solid  is  called  the  Hexaedron. 

The  angle  of  the  regular  pentagon  being  f  of  a  right  angle,  we 
have  |x3<[4;  but  |x4>4;  hence,  with  plane  angles  equal  to 
those  of  the  regular  pentagon,  we  can  form  only  a  triedral  angle. 
The  solid  bounded  by  twelve  regular  pentagons,  and  having  twenty 
solid  angles,  is  called  the  Dodecaedron. 

There  are,  then,  but  five  regular  polyedrons,  viz. :  The  tetraedron, 
the  octaedron,  and  the  icosaedron,  each  of  which  has  the  equilateral 
triangle  for  faces ;  the  hexaedron,  whose  faces  are  equal  squares, 
and  the  dodecaedron,  whose  faces  are  equal  regular  pentagons. 

It  is  obvious  that  a  sphere  may  be  circumscribed  about,  or  in- 
scribed within,  any  of  these  regular  solids,  and  conversely :  and 


SECTION  VI. 


381 


that  these  spheres  will  have  a  common  center,  which  may  also  be 
taken  as  the  center  of  the  polyedron. 

Any  regular  polyedron  may  be  regarded  as  made  up  of  a  number 
of  regular  pyramids,  whose  bases  are  severally  the  faces  of  the 
polyedron,  and  whose  common  vertex  is  its  center.  Each  of  these 
pyramids  will  have,  for  its  altitude,  the  radius  of  the  inscribed 
sphere;  and  since  the  volume  of  the  pyramid  is  measured  by  one 
third  of  the  product  of  its  base  and  altitude,  it  follows  that  the 
volume  of  any  regular  polyedron  is  measured  by  its  surface  multi- 
plied by  one  third  of  the  radius  of  the  inscribed  sphere. 


PROBLEM. 

Given,  the  name  of  a  regular  'polyedron,  and  the  side  of  the  hound- 
ing polygon,  to  find  the  inclination  of  its  faces;  the  radii  of  the  in- 
scribed and  circumscribed  spheres  ;  the  area  of  its  surface  ;  and  its 
volume. 

Let  AB  be  the  intersection  of  two  adjacent  faces  of  the  polye- 
dron, and  G  and  D  the  centers  of  these  faces,  0  being  the  center 
of  the  polyedron.  Draw  the  radii, 
OG  and  OD,  of  the  inscribed,  and 
]bhe  radii  OA  and  OB,of  the  circum- 
scribed sphere ;  also  from  G  and  D 
let  fall  the  perpendiculars  CE  and 
DE,  on  the  edge  AB,  and  draw  OE; 
then  will  the  angle  DEG  measure 
the  inclination  of  the  faces  of  the 
polyedron,  and  the  angle  DEO  is 
one  half  of  this  inclination. 

Let  i"  denote  the  inclination  of  the 
faces,  m  the  number  of  faces  which 
meet  to  form  a  polyedral  angle,  n  the 
number  of  sides  in  each  face,  and 
suppose  the  edge  of  the  polyedron  to 
be  unity. 

The  surface  of  the  sphere  of  which  0  is  the  center,  and  radius 
unity,  will  form,  by  its  intersections  with  the  planes,  AOE,  AOD, 
DOE,  the  right-angled  spherical  triangle  dae,  right-angled  at  e. 
In  the  right-angled  triangle  DEO,  the  angle  DOE  is  equal  to 


^oil  ua^j 


OF  THF 


382  SPHERICAL  TRIGONOMETRY. 

90°— -DEO  =  90°  —  \I, 

and  is  measured  by  the  are  de.     The  angle  due,  of  the  spherical 

*  •       i     •  i  *    360°       v..V  ,        ,        360° 

triangle,  is  equal  to ,  and  the  angle  ade  =  — — . 

2m  2n 

Now,  by  Napier's  Rules  we  have 

cos.c?ae  =  sin. ade  cos.de. 

-,         cos.dae  ,  -.  x 

or,  cos.de  =  _, ;  ( 1 ) 

sin. ade 

and,  cos.ae?  =  cot.dae  cot.ade  (2) 

Substituting  in  eq.  ( 1 ),  for  the  angles  dae  and  ade,  their  values, 

we  find 

cos.3600 

2m  /o\ 

Sin-2J=  ;sof-  [  ] 

2n    . 

Equation  (3  )  gives  the  value  of  the  sine  of  one  half  of  the  incli- 
nation of  the  planes  ;  and  by  means  of  this  equation  we  may  readily 
find  the  radii  of  the  inscribed  and  circumscribed  spheres. 

In  the  triangle  BED,  we  have 

DE  =  BE  cot.BDE  =  Jcot.  „, 

2n 

since  AB  =  1,  and  BE=  \AB. 
In  the  triangle  DOE,  we  have 

OD  =  DE  tan.  \1  =  Jcot.  _  tan. J 7  (4) 

2n 

From  the  triangle  A  OD,  we  find 

cos.DOA    :   1    ::    OD   :    OA 

whence  OA  = 

cos.DOA 

But  the  angle  DO  A  is  measured  by  the  arc  ad)  hence,  substi- 
tuting in  this  last  equation  the  values  of  cos.DOA  and  OD,  taken 
from  eqs.  (2)  and  (4),  we  have 

O4=itan.l/cot.???!  X  l X  —i— 

2  2n         cot360°        cot.  360° 

"ST  2w 

==  4  tan.  J  J  tan.  I!*!?!,  (5) 

2m 

by  writing  tan.  for  — ,  and  reducing, 
cot. 


SECTION  IV.  383 

Equation  ( 4 )  gives  the  value  of  OD,  the  radius  of  the  inscribed 
sphere,  and  equation  (5)  gives  that  of  OA,  the  radius  of  the  cir- 
cumscribed sphere.  The  area  of  one  of  the  faces  of  the  polyedron 
is  equal  to  one  half  of  the  apothegm  multiplied  by  the  perimeter. 

The  apothegm,  as  found  above,  is  equal  to  £  cot. ;  hence,  we 

2n 

360° 

have  JjixI  cot ,  for  the  area  of  one  of  the  faces;  and  multi- 

2n 

plying  this  by  the  number  of  faces  of  the  polyedron,  we  will  have 

the  expression  for  its  entire  area.     The  expression  for  the  surface 

multiplied  by  one  third  of  the  radius  of  the  inscribed  sphere,  gives 

the  measure  of  the  volume  of  the  polyedron. 

In  what  precedes,  we  have  supposed  the  edge  of  the  polyedron 
to  be  unity.  Having  found  the  radii  of  the  inscribed  and  circum- 
scribed spheres,  the  surfaces,  and  the  volumes  of  such  polyedrons, 
to  determine  the  radii,  surfaces,  and  volumes  of  regular  polyedrons 
having  any  edge  whatever,  we  have  merely  to  remember  that  the 
homologous  dimensions  of  similar  bodies  are  proportional;  their 
surfaces  are  as  the  squares  of  these  dimensions ;  and  their  volumes 
as  the  cubes  of  the  same. 

Formula  (3)  gives,  for  the  inclination  of  the  adjacent  faces  of 


The  Tetraedron,         70° 

31'  42" 

"    Hexaedron,         90° 

00'  00" 

"    Octaedron,         109° 

28/  18" 

"    Dodecaedron,    116° 

33'  54" 

"    Icosaedron,       138° 

11'  23" 

The  subjoined  table  gives 

the  surfaces  and  volumes  of  the  regular 

polyedrons,  when  the  edge  is  unity. 

Surfaces. 

Volumes. 

Tetraedron, 

1.7320508 

0.1178513 

Hexaedron, 

6.0000000 

1.0000000 

Octaedron, 

3.4641016 

0.4714045 

Dodecaedron, 

20.6457288 

7.6631189 

Icosaedron, 

8.6602540 

2.1816950 

LOGARITHMIC  TABLES; 


ALSO    A    TABLE    OF 


NATURAL    AND     LOGARITHMIC 


SINES,  COSINES,  AND  TANGENTS, 


TO   EVERY   MINUTE    OF   TEE    QUADRANT. 


LOGARITHMS    OF 

NUMBERS 

FROM 

1    to    lOOOO, 

N. 

Log. 

N. 

Log. 

N. 

Log. 

N. 

Log. 

1 

0  000000 

26 

1  414973 

51 

1  707570 

76 

1  880814 

2 

0  301030 

27 

1  431364 

52 

1  716003 

77 

1  888491 

3 

0  477121 

28 

1  447158 

53 

1  724276 

78 

1  892095 

4 

0  602030 

29 

1  462398 

54 

1  732394 

79 

1  897627 

5 

0  698970 

30 

1  477121 

55 

1  740363 

80 

1  903090 

6 

0  778151 

31 

1  491362 

56  . 

1  748188 

81 

1  908485 

7 

0  845098 

32 

1  505150 

57 

1  755875 

82 

1  913814 

8 

0  903090 

33 

1  518514 

58 

1  763428 

83 

1  919078 

9 

0  954243 

34 

1  531479 

59 

1  770852 

84 

1  924279 

10 

1  000000 

35 

1  544068 

60 

1  778151 

85 

1  929419 

11 

1  041393 

36 

1  556303 

61 

1  785330 

86 

1  934498 

12 

1  079181 

37 

1  568202 

62 

1  792392 

87 

1  939519 

13 

1  113943 

38 

1  579784 

63 

1  799341 

88 

1  944483 

14 

1  146128 

39 

1  591035 

64 

1  803180 

89 

1  949390 

15 

1  176091 

40 

1  602030 

65 

1  812913 

90 

1  954243 

16 

1  204120 

41 

1  612784 

66 

1  819544 

91 

1  959041 

17 

1  230449 

42 

1  623249 

67 

1  826075 

92 

1  963788 

18 

1  255273 

43 

1  633468 

68 

1  832509 

93 

1  968483 

19 

1  278754 

44 

1  643453 

69 

1  838849 

94 

1  9/3128 

20 

1  301030 

45 

1  653213 

70 

1  845098 

95 

1  977724 

21 

1  322219 

46 

1  662578 

71 

1  851258 

96 

1  982271 

22 

1  342423 

47 

1  672098 

72 

1  857333 

97 

1  986772 

23 

1  361728 

48 

1  681241 

73 

1  863323 

98 

1  991226 

24 

1  380211 

49 

1  690196 

74 

1  869232 

99 

1  995635 

25 

1  397940 

50 

1  698970 

75 

1  875081 

100 

2  000000 

Si 

)TE.  In  the  following  table,  in  the  last  ni 

he  columns  of  each  p 

age,  where 

the 

irst  or  leading  figures  change  from  9's 

»  to  0's,  points  or  do 

ts  are  now 

intri 

jduced  instead  of  the  0's  through  the  r 

sst  of  the  line,  to  cat< 

ih  the  eye, 

and 

to  indicate  that  from  thence  the  corr 

ssponding  natural    r 

mmber   in 

the 

irst  column  stands  in  the  next  lower 

line,  and  its  annexe 

i  first  two 

figu 

res  of  the  Logarithms  in  the  second  co 

lumn. 

N.  ; 

LOGARITHMS    OF     NUMBERS.               3 

0 

1 
0434 

0S68 

3 

4 

5 

6 

7 

8 

9 

100 

000000 

1301 

1734 

2166 

2598 

3029 

3461 

3891 

101 

4321 

4750 

5181 

5609 

6038 

6468 

6894 

7321 

7748 

8174 

102 

8600 

9026 

9451 

9876 

.300 

.724 

1147 

1570 

1993 

2415 

j 

103 

012837 

3259 

3680 

4100 

4521 

4940 

5360 

5779 

6197 

6616 

104 

7033 

7451 

7888 

8284 

8700 

9116 

9532 

9947 

.361 

.775 

105 

021189 

1603 

2016 

2428 

2841 

3252 

3664 

4075 

4486 

4896 

103 

530S 

5715 

6125 

6533 

6942 

7350 

7757 

8164 

8571 

8978 

107 

9384 

9789 

.195 

.600 

1004 

1408 

1812 

2216 

£619 

3021 

108 

033424 

3826 

4227 

4628 

5029 

5430 

5830 

6230 

6629 

7028 

109 

7426 

7825 

8223 

8620 

9017 

9414 

9811 

.207 

.602 

.998 

110 

041393 

1787 

2182 

2576 

2969 

3362 

3755 

4148 

4540 

4932 

111 

5323 

5714 

6105 

6495 

6885 

7275 

7664 

8053 

8442 

8830 

112 

9218 

9608 

9993 

.380 

.766 

1153 

1538 

1924 

2309 

2694 

113 

053078 

3483 

3846 

4230 

4813 

4996 

5378 

5760 

6142 

6524 

114 

6905 

7286 

7666 

8046 

8426 

8805 

9185 

9563 

9942 

.320 

115 

030398 

1075 

1452 

1829 

2206 

2582 

2958 

3333 

3709 

4083 

116 

4458 

4832 

5203 

5580 

5953 

6326 

6699 

7071 

7443 

7815 

117 

8186 

8557 

8928 

9298 

9668 

..38 

.407 

.776 

1145 

1514 

118 

071882 

2250 

2617 

2985 

3352 

3718 

4085 

4451 

4816 

5182 

119 

5547 

5912 

6276 

6640 

7004 

7368 

7731 

8094 

8457 

8819 

120 

9181 

9543 

9904 

.266 

.626 

.987 

1347 

1707 

2067 

2426 

121 

082785 

3144 

3503 

3861 

4219 

4576 

4934 

5291 

5647 

6004 

122 

6360 

6716 

7071 

7426 

7781 

8136 

8490 

8845 

9198 

9552 

123 

9905 

.258 

.611 

.963 

1315 

1667 

2018 

2370 

2721 

3071 

124 

093422 

3772 

4122 

4471 

4820 

5169 

5518 

5866 

6215 

6562 

125 

6910 

7257 

7604 

7951 

8298 

8644 

8990 

9335 

9681 

1026 

126 

100371 

0715 

1059 

1403 

1747 

2091 

2434 

2777 

3119 

3462 

127 

3804 

4146 

4487 

4828 

5169 

5510 

5851 

6191 

6531 

6871 

128 

7210 

7549 

7S88 

8227 

8565 

8903 

9241 

9579 

9916 

.253 

129 

110590 

0926 

1263 

1599 

1934 

2270 

2605 

2940 

3275 

3609 

130 

3943 

4277 

4611 

4944 

5278 

5611 

5943 

6276 

6608 

6940 

131 

7271 

7603 

7934 

8265 

8595 

8926 

9256 

9586 

9915 

0245 

132 

120574 

0903 

1231 

1560 

1888 

2216 

2544 

2871 

3198 

3525 

133 

3852 

4178 

4504 

4830 

5158 

5481 

5806 

6131 

6456 

6781 

134 

7105 

7429 

7753 

8076 

8399 

8722 

9045 

9368 

9690 

..12 

135 

130334 

0655 

0977 

1298 

1619 

1939 

2260 

2580 

2900 

3219 

136 

3539 

3858 

4177 

4496 

4814 

5133 

5451 

5769 

6086 

6403 

137 

6721 

7037 

7354 

7671 

7987 

8303 

8618 

8934 

9249 

9564 

138 

9879 

.194 

.508 

.822 

1136 

1450 

1763 

2076 

2389 

2702 

1 

139 

143015 

3327 

3630 

3951 

4263 

4574 

4885 

5196 

5507 

5818 

140 

6128 

6438 

6748 

7058 

7367 

7676 

7985 

8294 

8603 

8911 

141 

9219 

9527 

9835 

.142 

.449 

.756 

1063 

1370 

1676 

1982 

142 

152288 

2594 

2900 

£205 

3510 

3815 

4120 

4424 

4728 

5032 

143 

5336 

5640 

5943 

6246 

6549 

6852 

7154 

7457 

7759 

8061 

144 

8362 

8664 

8965 

9266 

9567 

9868 

.168 

.469 

.769 

1068 

145 

161368 

1667 

1967 

2266 

2564 

2863 

3161 

3460 

3758 

4055 

146 

4353 

4650 

4947 

5244 

5541 

5838 

6134 

6430 

6726 

7022 

147 

7317 

7613 

7908 

8203 

8497 

8792 

9086 

9380 

9674 

9968 

148 

170262 

0555 

0848 

1141 

1434 

1726 

2019 

2311 

2603 

2895 

149 

3186 

3478 

3769 

4060 

4351 

4641 

4932 

5222 

5512 

5802 

1* 


4 

LOGARITHMS 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

150 

176091 

6381 

6670 

6959 

7248 

7536 

7825 

8113 

8401 

8689 

151 

8977 

9264 

9552 

9839 

.126 

.413 

.699 

.985 

1272 

1558 

152 

181844 

2129 

2415 

2700 

2985 

3270 

3555 

3839 

4123 

4407 

153 

4691 

4975 

5259 

5542 

5825 

6108 

6391 

6674 

6956 

7239 

154 

7521 

7803 

8084 

8366 

8647 

281 

1451 

8928 

9209 

9490 

9771 

..51 

155 

190332 

0812 

0892 

1171 

1730 

2010 

2289 

2567 

2846 

156 

3125 

3403 

3681 

3959 

4237 

4514 

4792 

5069 

5346 

5623 

157 

5899 

6176 

6453 

6729 

7005 

7281 

7556 

7832 

8107 

8382 

158 

8657 

8932 

9206 

9481 

9755 

..29 

.303 

.577 

.850 

1124 

159 

201397 

1670 

1943 

2216 

2488 
273 

2761 

3033 

3305 

3577 

3848 

160 

4120 

4391 

4663 

4934 

5204 

5475 

5746 

6016 

6286 

6556 

161 

6826 

7096 

7365 

7634 

7904 

8173 

8441 

8710 

8979 

9247 

162 

9515 

9783 

..51 

.319 

.586 

.853 

1121 

1388 

1654 

1921 

163 

212188 

2454 

2720 

2986 

3252 

3518 

3783 

4049 

4314 

4579 

164 

4844 

5109 

5373 

5638 

5902 
264 

6166 

6430 

6694 

6957 

7221 

165 

7484 

7747 

8010 

8273 

8536 

8798 

9060 

9323 

9585 

9846 

166 

220108 

0370 

0831 

0892 

1153 

1414 

1675 

1936 

2196 

2456 

167 

2716 

2976 

3236 

3496 

3755 

4015 

4274 

4533 

4792 

5051 

168 

5309 

5568 

5fe26 

6084 

6342 

6600 

6858 

7115 

7372 

7630 

169 

7887 

8144 

8400 

8657 

8913 

257 

9170 

9426 

9682 

9938 

.193 

170 

230449 

0704 

0960 

1215 

1470 

1724 

1979 

2234 

2488 

2742 

171 

2996 

3250 

3504 

3757 

4011 

4264 

4517 

4770 

5023 

5276 

172 

5528 

5781 

6033 

6285 

6537 

6789 

7041 

7292 

7544 

7795 

173 

8046 

8297 

8548 

8799 

9049 

9299 

9550 

9800 

..50 

.300 

174 

240549 

0799 

1048 

1297 

1546 
249 

1795 

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OF  NUMBERS.              7 

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5838 

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6078 

6157 

6237 

6317 

545 

6397 

6476 

6556 

6636 

6795 

6874 

6954 

7034 

7113 

546 

7193 

7272 

7352 

7431 

7511 

7590 

7670 

7749 

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517 

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8146 

8225 

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8463 

8543 

8622 

8701 

548 

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8939 

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9177 

9256 

9335 

9414 

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549 

9572 

9651 

9731 

9810 

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9968 

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.205 

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12 

LOGARITHMS 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

550 

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0560 

0678 

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0836 

0915 

0994 

1073 

551 

1152 

1230 

1309 

1388 

1467 

1546 

1624 

1703 

1782 

1860 

552 

1939 

2018 

2096 

2175 

2254 

2332 

2411 

2489 

2568 

2646 

553 

2725 

2804 

2882 

2961 

3039 

3118 

3196 

3275 

3353 

3431 

554 

3510 

3558 

3667 

3745 

3823 

19 

4608 

3902 

3980 

4058 

4136 

4215 

555 

4293 

4371 

4449 

4528 

4684 

4762 

4840 

4919 

4997 

556 

5075 

5153 

5231 

5309 

5387 

5465 

5543 

5621 

5699 

5777 

557 

5855 

5933 

6011 

6089 

6167 

6245 

6323 

6401 

6479 

6556 

558 

6634 

6712 

6790 

6868 

6945 

7023 

7101 

7179 

7256 

7334 

559 

7412 

7489 

7567 

7645 

7722 

7800 

7878 

7955 

8033 

8110 

560 

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8266 

8343 

8421 

8498 

8576 

8653 

8731 

8808 

8885 

561 

8963 

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9350 

9427 

9504 

9582 

9659 

562 

9736 

9814 

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9968 

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.277 

.354 

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563 

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0586 

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0740 

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0971 

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1125 

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564 

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1433 

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1587 

1664 

1741 

1818 

1895 

1972 

565 

2048 

2125 

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2279 

2356 

2433 

2509 

2586 

2663 

2740 

566 

2816 

2893 

2970 

3047 

3123 

3200 

3277 

3353 

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567 

3582 

3660 

3736 

3813 

3889 

3966 

4042 

4119 

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4272 

568 

4348 

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4960 

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569 

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5722 

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570 

5875 

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6027 

6103 

61F0 

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6332 

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6484 

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571 

6636 

6712 

6788 

6864 

6940 

7016 

7092 

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7244 

7320 

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7396 

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7700 

7775 

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7927 

8003 

8079 

573 

8155 

8230 

8308 

8382 

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8533 

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8761 

8836 

574 

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8988 

9068 

9139 

9214 

74 

9970 

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9366 

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9517 

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575 

9638 

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9819 

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..45 

.121 

.196 

.272 

.347 

576 

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0573 

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0724 

0799 

0875 

0950 

1025 

1101 

577 

1176 

1251 

1326 

1402 

1477 

1552 

1627 

1702 

1778 

1853 

578 

1923 

2003 

2078 

2153 

2228 

2303 

2378 

2453 

2529 

2604 

579 

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2754 

2829 

2904 

2978 

3053 

3128 

2203 

3278 

3353 

580 

3428 

3503 

3578 

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3727 

3802 

3877 

3952 

4027 

4101 

581 

4176 

4251 

4326 

4400 

4475 

4550 

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582 

4923 

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5072 

5147 

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5445 

5520 

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583 

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5743 

5818 

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5966 

6041 

6115 

6190 

6264 

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584 

6413 

6487 

6562 

6636 

6710 

6785 

6859 

6933 

7007 

7082 

585 

7156 

7230 

7304 

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7527 

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586 

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7972 

8046 

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8268 

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9230 

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588 

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9746 

9820 

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589 

770115 

0189 

0263 

0336 

0410 

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0557 

0631 

0705 

0778 

590 

0852 

0926 

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1073 

1146 

1220 

1293 

1367 

1440 

1514 

591 

1587 

1661 

1734 

1808 

1881 

1955 

2028 

2102 

2175 

2248 

592 

2322 

2395 

2468 

3542 

2615 

2688 

2762 

2835 

2908 

2981 

593 

3055 

3128 

3201 

3274 

3348 

3421 

3494 

3567 

3640 

3713 

594 

3786 

3860 

3933 

4006 

4079 
73 

4809 

4152 

4225 

4298 

4371 

4444 

595 

4517 

4590 

1663 

4736 

4882 

4955 

5028 

5100 

6173 

596 

5246 

5319 

5392 

5465 

5538 

5610 

5683 

5756 

5829 

5902 

597 

5974 

6047 

61120 

6193 

6265 

6338 

6411 

6483 

6556 

6629 

598 

6701 

6774 

6846 

6919 

6992 

7064 

7137 

7209 

7282 

7354 

599 

7427 

7499 

7572 

7644 

7717 

7789 

7862 

7934 

8006 

8079 

OF  NUMBERS.              13 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

600 

778151 

8224 

8296 

8368 

8441 

8513 

8585 

8658 

8730 

8802 

601 

8874 

8947 

9019 

9091 

9163 

9236 

9308 

9380 

9452 

9524 

602 

9596 

6669 

9741 

9813 

9885 

9957 

..29 

.101 

.173 

.245 

603 

780317 

0389 

0461 

0533 

0605 

0677 

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0893 

0965 

604 

1037 

1109 

1181 

1253 

1324 

n 

2042 

1396 

1468 

1540 

1612 

1684 

605 

1755 

1827 

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1971 

2114 

2186 

2258 

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2401 

606 

2473 

2544 

2616 

2688 

2759 

2831 

2902 

2974 

3046 

3117 

607 

3189 

3260 

3332 

3403 

3475 

3546 

3618 

3689 

3761 

3832 

608 

3904 

3975 

4046 

4118 

4189 

4261 

4332 

4403 

4475 

4546 

609 

4617 

4689 

4760 

4831 

4902 

4974 

5045 

5116 

5187 

5259 

610 

5330 

5401 

5472 

5543 

5615 

5686 

5757 

5828 

5899 

5970 

611 

6041 

6112 

6183 

6254 

6325 

6396 

6467 

6538 

6609 

6680 

612 

6751 

6822 

6893 

6964 

7035 

7106 

7177 

7248 

7319 

7390 

613 

7460 

7531 

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7744 

7815 

7885 

79^6 

8027 

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614 

8168 

8239 

8310 

8381 

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8522 

8593 

8663 

8734 

8804 

615 

8875 

8946 

9016 

9087 

9157 

9228 

9299 

9369 

9440 

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616 

9581 

9651 

9722 

9792 

9863 

9933 

...4 

..74 

.144 

.215 

617 

790285 

0356 

0426 

0496 

0567 

0637 

0707 

0778 

0848 

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618 

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1059 

1129 

1199 

1269 

1340 

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1480 

1550 

1620 

619 

1691 

1761 

1831 

1901 

1971 

2041 

2111 

2181 

2252 

2322 

620 

2392 

2462 

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2602 

2672 

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2952 

3022 

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3162 

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3301 

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3860 

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4000 

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4139 

4209 

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4349 

4418 

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4488 

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4627 

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4767 

4836 

4906 

4976 

5045 

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624 

5185 

5254 

5324 

5393 

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69 

6158 

5532 

5602 

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5741 

5811 

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625 

5880 

5949 

6019 

6088 

6227 

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6366 

6436 

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626 

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6713 

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6852 

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7060 

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627 

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7683 

7752 

7821 

7890 

628 

7960 

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8167 

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8374 

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8513 

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629 

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8720 

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8858 

8927 

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9065 

6134 

9203 

9272 

630 

9341 

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9547 

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9823 

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9961 

631 

800026 

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0373 

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; 

632 

0717 

0786 

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0923 

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1061 

1129 

1198 

1266 

1335 

633 

1404 

1472 

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1609 

1678 

1747 

1815 

1884 

1952 

2021 

634 

2089 

2158 

2226 

2295 

2363 

2432 

2500 

2568 

2637 

2705 

635 

2774 

2842 

2910 

2979 

3047 

3116 

3184 

3252 

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3389. 

636 

3457 

3525 

3594 

3662 

3730 

3798 

3867 

3935 

4003 

4071 

637 

4139 

4208 

4276 

4354 

4412 

4480 

4548 

4616 

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4753 

638 

4821 

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4957 

5025 

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5229 

5297 

5365 

5433 

639 

5501 

5669 

5637 

5705 

5773 

5841 

5908 

5976 

6044 

6112 

640 

6180 

6248 

6316 

6384 

6451 

6519 

6587 

6655 

6723 

6790 

641 

6858 

6926 

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7061 

7129 

7157 

7264 

7332 

7400 

7467 

642 

7535 

7603 

7670 

7738 

7806 

7873 

7941 

8008 

8076 

8143 

643 

8211 

8279 

8346 

8414 

8481 

8549 

8616 

8684 

8751 

8818 

644 

8886 

8953 

9021 

9088 

9156 

9223 

9290 

9358 

9425 

9492 

645 

9560 

9627 

9694 

9762 

9829 

9896 

9964 

..31 

..98 

.165 

646 

810233 

0300 

0367 

0434 

0501 

0596 

0636 

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0770 

0837 

647 

0904 

0971 

1039 

1106 

1173 

1240 

1307 

1374 

1441 

1508 

648 

1575 

1642 

1709 

1776 

1843 

1910 

1977 

2044 

2111 

2178  1 

649 

2245 

2312 

2379 

2445 

2512 

2579 

2646 

2713 

2780 

2847 

14 

LOGARITHMS 

N. 
650 

0 
812913 

1 

o 

3 

4 

5 

6 

7 

8 

9 

2980 

3047 

3114 

3181 

3247 

3314 

3381 

3448 

3514 

651 

3581 

3648 

3714 

3781 

3848 

3914 

3981 

4048 

4114 

4181 

652 

4248 

4314 

4381 

4447 

4514 

4581 

4647 

4714 

4780 

4847 

653 

4913 

4980 

5046 

5113 

5179 

5246 

5312 

5378 

5445 

5511 

654 

5578 

5644 

5711 

5777 

5843 

67 

6508 

5910 

5976 

6042 

6109 

6175 

655 

6241 

6308 

6374 

6440 

6573 

6639 

6705 

6771 

6838 

656 

6904 

6970 

7036 

7102 

7169 

7233 

7301 

7367 

7433 

7499 

657 

7565 

7631 

7698 

7764 

7830 

7896 

7962 

8028 

8094 

8160 

658 

8226 

8292 

8358 

8424 

8490 

8556 

8622 

8688 

8754 

8820 

659 

8885 

8951 

9017 

9083 

9149 

9215 

9281 

9346 

9412 

9478 

660 

9544 

9610 

9676 

9741 

9807 

9873 

9939 

...4 

..70 

.136 

661 

820201 

0267 

0333 

0399 

0464 

0530 

0595 

0661 

0727 

0792 

662 

0858 

0924 

0939 

1055 

1120 

1186 

1251 

1317 

1382 

1448 

663 

1514 

1579 

1645 

1710 

1775 

1841 

1906 

1972 

2037 

2103 

664 

2168 

2233 

2299 

2364 

2430 

2495 

2560 

2626 

2691 

2756 

665 

2822 

2887 

2952 

3018 

3083 

3148 

3213 

3279 

3344 

3409 

666 

3474 

3539 

3605 

3670 

3735 

3800 

3865 

3930 

3996 

4081 

667 

4126 

4191 

4256 

4321 

4386 

4451 

4516 

4581 

4646 

4711 

668 

4776 

4841 

4906 

4971 

5036 

5101 

5166 

5231 

5296 

5361 

669 

5426 

5491 

5556 

5621 

5686 

5751 

5815 

5880 

5945 

6010 

670 

6075 

6140 

6204 

6269 

6334 

6399 

6464 

6528 

6593 

6658 

671 

6723 

6787 

6852 

6917 

6981 

7046 

7111 

7175 

7240 

7305 

672 

7369 

7434 

7499 

7563 

7628 

7692 

7757 

7821 

7886 

7951 

673 

8015 

8080 

8144 

8209 

8273 

8338 

8402 

8467 

8531 

8595 

674 

8660 

8724 

8789 

8853 

8918 

65 

9561 

8982 

9046 

9111 

9175 

9239 

675 

9304 

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9432 

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9625 

9690 

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9818 

9882 

676 

9947 

..11 

..75 

.139 

.204 

.268 

.332 

.396 

.460 

.525 

677 

830589 

0853 

0717 

0781 

0845 

0909 

0973 

1037 

1102 

1166 

678 

1230 

1294 

1358 

1422 

1486 

1550 

1614 

1678 

1742 

1808 

679 

1870 

1934 

1998 

2062 

2126 

2189 

2253 

2317 

2381 

2445 

680 

2509 

2573 

2637 

2700 

2764 

2828 

2892 

2956 

3020 

3083 

681 

3147 

3211 

3275 

3338 

3402 

3466 

3530 

3593 

3657 

3721 

682 

3784 

3848 

3912 

3975 

4039 

4103 

4166 

4230 

4294 

4357 

683 

4421 

4484 

4548 

4611 

4675 

4739 

4802 

4866 

4929 

4993 

684 

5056 

5120 

5183 

5247 

5310 

5373 

5437 

5500 

5564 

5627 

685 

5691 

5754 

5817 

5881 

5944 

6007 

6071 

6134 

6197 

6261 

686 

6324 

6387 

6451 

6514 

6577 

6641 

6704 

6767 

6830 

6894 

687 

6957 

7020 

7083 

7146 

7210 

7273 

7336 

7399 

7462 

7525 

688 

7588 

7652 

7715 

7778 

7841 

7904 

7967 

8030 

8093 

8156 

689 

8219 

8282 

8345 

8408 

8471 

8534 

8597 

8660 

8723 

8786 

690 

8849 

8912 

8975 

9038 

9109 

9164 

9227 

9289 

9352 

9415 

691 

9478 

9541  9604 

9667 

9729 

9792 

9855 

9918 

9981 

..43 

692 

840106 

0169  0232 

0294 

0357 

0420 

0482 

0545 

0608 

0671 

693 

0733 

0796  0859 

0921 

0984 

1046 

1109 

1172 

1234 

1297 

694 

1359 

1422  1485 

1547 

1610 

62 
2235 

1672 

1735 

1797 

1860 

1922 

695 

1985 

2047  2110 

2172 

2297 

2360 

2422 

2484 

2547 

696 

2609 

2672  2734 

2796 

2859 

2921 

2983 

3046 

3108 

3170 

697 

3233 

3295  3357 

3420. 

3482 

3544 

3606 

3669 

3731 

3/93 

698 

3855 

3918  3980 

4042 

4104 

4166 

4229 

4291 

4353 

4415 

699 

4477 

4539  4601 

4664 

4726 

4788 

4850 

4912 

4974 

5036 

OF  NU  MBERS.             15 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

700 

845098 

5160 

5222 

5284 

5346 

5408 

5470 

5532 

5594 

5656 

701 

5718 

5780 

5842 

5904 

5966 

6028 

6090 

6151 

6213 

6275 

702 

6337 

6399 

6461 

6523 

6585 

6646 

6708 

6770 

6832 

6894 

703 

6955 

7017 

70/9 

7141 

7202 

7264 

7326 

7388 

7449 

7511 

704 

7573 

7634 

7676 

7758 

7819 

62 

8435 

7831 

7943 

8004 

8066 

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705 

8189 

8251 

8312 

8374 

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8559 

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8743 

70S 

8805 

8866 

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9051 

9112 

9174 

9235 

9297 

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707 

9419 

9481 

9542 

9604 

9665 

9726 

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708 

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0095 

0156 

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0279 

0340 

0401 

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709 

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0707 

0769 

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0952 

1014 

1075 

1136 

1197 

710 

1258 

1320 

1381 

1442 

1503 

1564 

1625 

1686 

1747 

1809 

711 

1870 

1931 

1992 

2053 

2114 

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2236 

2297 

2358 

2419 

712 

2480 

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2602 

2663 

2724 

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2846 

2907 

2968 

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713 

3090 

3150 

3211 

3272 

3333 

3394 

3455 

3516 

3577 

3637 

714 

3698 

3759 

3820 

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3941 

4002 

4063 

4124 

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715 

4308 

4367 

4428 

4488 

4549 

4610 

4670 

4731 

4792 

4852 

716 

4913 

4974 

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5216 

5277 

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5398 

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717 

5519 

5580 

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6003 

6064 

718 

6124 

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6306 

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6427 

6487 

6548 

6608 

6668 

719 

6729 

6789 

6850 

6910 

6970 

7031 

7091 

7152 

7212 

7272 

720 

7332 

7393 

7453 

7513 

7574 

7634 

7694 

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7815 

7875 

721 

7935 

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8056 

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8176 

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8357 

8417 

8477 

722 

8537 

8597 

8657 

8718 

8778 

8838 

8898 

8958 

9018 

9078 

723 

9138 

9198 

9258 

9318 

9379 

9439 

9499 

9559 

9619 

9679 

724 

9739 

9799 

9859 

9918 

9978 

60 

0578 

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..98 

.158 

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725 

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0398 

0458 

0518 

0637 

0697 

0757 

0817 

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720 

0937 

0996 

1056 

1116 

1176 

1236 

1295 

1355 

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1475 

727 

1534 

1594 

1654 

1714 

1773 

1833 

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1952 

2012 

2072 

728 

2131 

2191 

2251 

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2370 

2430 

2489 

2549 

2608 

2668 

729 

2728 

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2847 

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3501 

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3620 

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3739 

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3917 

3977 

4036 

4096 

4155 

4214 

4274 

4333 

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4511 

4570 

4630 

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4148 

4808 

4867 

4926 

4985 

5045 

733 

5104 

5163 

5222 

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5400 

5459 

5519 

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5637 

734 

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5755 

5814 

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5933 

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6051 

6110 

6169 

6228 

735 

6287 

6346 

6405 

6465 

6524 

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6701 

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1047 

1106 

1164 

1223 

1281 

1339 

1398 

1456 

1515 

744 

1573 

1631 

1690 

1748 

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2389 

1865 

1923 

1981 

2040 

2098 

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2156 

2215 

2273 

2331 

2448 

2506 

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2622 

2681 

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2739 

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2913 

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3727 

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3844 

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3902 

3960 

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4192 

4250 

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4360 

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4540 

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4830 

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4945 

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16 

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N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

750 

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5235 

5293 

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5409 

5466 

5524 

5582 

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5640 

5698 

5756 

5813 

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5929 

5987 

6045 

6102 

6160 

752 

6218 

6276 

6333 

6391 

6449 

6507 

6564 

6622 

6680 

6737 

753 

6795 

6853 

6910 

6968 

7026 

7083 

7141 

7199 

7256 

7314 

754 

7371 

7429 

7487 

7544 

7602 

57 

8177 

7659 

7717 

7774 

7832 

7889 

755 

7947 

8004 

8062 

8119 

8234 

8292 

8349 

8407 

8464 

756 

8522 

8579 

8637 

8694 

8752 

8809 

8866 

8924 

8981 

9039 

757 

9096 

9153 

9211 

9268 

9325 

9383 

9440 

9497 

9555 

9612 

758 

9669 

9726 

9784 

9841 

9898 

9956 

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..70 

.127 

.185 

759 

880242 

0299 

0356 

0413 

0471 

0528 

0580 

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0699 

0756 

760 

0814 

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0928 

0985 

1042 

1099 

1156 

1213 

1271 

1328 

761 

1385 

1442 

1499 

1556 

1613 

1670 

1727 

1784 

1841 

1898 

762 

1955 

2012 

2069 

2126 

2183 

2240 

2297 

2354 

2411 

2468 

763 

2525 

2581 

2638 

2695 

2752 

2809 

2866 

2923 

2980 

3037 

764 

3093 

3150 

3207 

3264 

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3377 

3434 

3491 

3548 

3605 

765 

3661 

3718 

3775 

3832 

3888 

3945 

4002 

4059 

4115 

4172 

766 

4229 

4285 

4342 

4399 

4455 

4512 

4569 

4625 

4682 

4739 

767 

4795 

4852 

4909 

4965 

5022 

5078 

5135 

5192 

5248 

5305 

768 

5361 

5418 

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5531 

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6700 

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5813 

5870 

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5926 

5983 

6039 

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6152 

6209 

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6378 

6434 

770 

6491 

6547 

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6660 

6716 

6773 

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6942 

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771 

7054 

7111 

7167 

7233 

7280 

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7617 

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773 

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777 

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778 

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1147 

1203 

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1314 

1370 

1426 

1482 

779 

1537 

1593 

1649 

1705 

1760 

1816 

1872 

1928 

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2039 

780 

2095 

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2208 

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2373 

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2540 

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2651 

2707 

2762 

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2873 

2929 

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3040 

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3318 

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3429 

3484 

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3595 

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3706 

783 

3762 

3817 

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3984 

4039 

4094 

4150 

4205 

4261 

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4371 

4427 

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4704 

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4814 

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4870 

4925 

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6140 

6195 

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6306 

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6912 

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7022 

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7132 

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7297 

7352 

7407 

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790 

7627 

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7737 

7792 

7847 

7902 

7957 

8012 

8067 

8122 

791 

8176 

8231 

8286 

8341 

8396 

8451 

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8561 

8615 

8670 

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8725 

8780 

8835 

8890 

8944 

8999 

9054 

9109 

9164 

9218 

793 

9273 

9328 

9383 

9437 

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9547 

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9711 

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794 

9821 

9875 

9930 

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795 

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0422 

0476 

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0804 

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796 

0913 

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1022 

1077 

1131 

1186 

1240 

1295 

1349 

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1458 

1513 

1567 

1622 

1676 

1736 

1785 

1840 

18*4 

1948 

798 

2003 

2057 

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2166 

2221 

2275 

2329 

2384 

2438 

2492 

799 

2547 

2601 

2655 

2710 

2764 

2818 

2873 

2927 

2981 

3036 

OF  NUMBERS 

17 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

800 

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3144 

3199 

3253 

3307 

3361 

3416 

3470 

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3633 

3687 

3741 

3795 

3849 

3904 

3958 

4012 

4066 

4120 

802 

4174 

4229 

4283 

4337 

4391 

4445 

4499 

4553 

4607 

4601 

803 

4716 

4770 

4824 

4878 

4932 

4986 

5040 

5094 

5148 

5202 

804 

5256 

5310 

5364 

5418 

5472 

54 

6012 

5526 

5580 

5634 

5688 

5742 

805 

6796 

5850 

5904 

5958 

6086 

6119 

6173 

6227 

6281 

80-3 

6335 

6389 

6443 

6497 

6551 

6604 

6658 

6712 

6766 

6820 

807 

6874 

6927 

6981 

7035 

7089 

7143 

7196 

7250 

7304 

7358 

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7411 

7465 

7519 

7573 

7626 

7680 

7734 

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7841 

7895 

809 

7949 

8002 

8058 

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8163 

8217 

8270 

8324 

8378 

8431 

810 

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8539 

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8860 

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1051 

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1211 

1264 

1317 

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1424 

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1637 

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1690 

1743 

1797 

1850 

1903 

1956 

2009 

2063 

2115 

2169 

817 

2222 

2275 

2323 

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2435 

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2541 

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2700 

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2753 

2808 

2859 

2913 

2966 

3019 

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3231 

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3284 

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3443 

3496 

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3602 

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3708 

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3814 

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3920 

3973 

4026 

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4132 

4184 

4237 

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9340 

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9914 

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920123 

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833 

0645 

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1010 

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1114 

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1166 

1218 

1270 

1322 

1374 

1426 

1478 

1530 

1582 

1634 

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1686 

1738 

1790 

1842 

1894 

1946 

1998 

2050 

2102 

2154 

836 

2208 

2258 

2310 

2362 

2414 

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3296 

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4021 

4072 

4124 

4147 

4228 

840 

4279 

4331 

4383 

4434 

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4538 

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4693 

4744 

841 

4796 

4848 

4899 

4951 

5003 

5054 

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5209 

6261 

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5312 

5364 

5415 

5467 

5518 

5570 

5621 

5073 

5725 

5776 

843 

5828 

5874 

5931 

5982 

6034 

6085 

6137 

6188 

6240 

6291 

844 

6342 

6394 

6445 

6497 

6548 
52 

7082 

6600 

6651 

6702 

6754 

6805 

845 

6857 

6908 

6959 

7011 

7114 

7165 

7216 

7268 

7319 

846 

7370 

7422 

7473 

7524 

7576 

7627 

7678 

7730 

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7935 

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8037 

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8242 

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8857 

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9031 

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9216 

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LOGARITHMS 

N. 

0 

I     2 

3   1 

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5 

6 

7 

8 

9 

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9725 

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9930 

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852 

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0745 

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0847 

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853 

0949 

1000 

1051 

1102 

1153 

1204 

1254 

1305 

1356 

1407 

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1458 

1509 

1560 

1610 

1661 

51 

2169 

1712 

1763 

1814 

1865 

1915 

855 

1966 

2017 

2068 

2118 

2220 

2271 

2322 

2372 

2423 

856 

2474 

2524 

2575 

2626 

2677 

2727 

2778 

2829 

2879 

2930 

857 

2981 

3031  3082 

3133 

3183 

3234 

3285 

3335 

3386 

3437 

853 

3487 

3538  3589 

3639 

3690 

3740 

3791 

3841 

3892 

3943 

859 

3993 

4044 

4094 

4145 

4195 

4246 

4269 

4347 

4397 

4448 

860 

4498 

4549 

4599 

4650 

4700 

4751 

4801 

4852 

4902 

4953 

861 

5003 

5054 

5104 

5154 

5205 

5255 

5306 

5356 

5406 

5457 

862 

5507 

5558 

5603 

5658 

5709 

5759 

5809 

5380 

5910 

5960 

863 

6011 

6051 

6111 

6162 

6212 

6262 

6313 

6363 

6413 

6463 

864 

6514 

6564 

6614 

6665 

6715 

6765 

6815 

6865 

6916 

6966 

865 

7016 

7056 

7117 

7167 

7217 

7267 

7317 

7367 

7418 

7468 

866 

7518 

7568 

7618 

7668 

7718 

7769 

7819 

7869 

7919 

7969 

867 

8019 

8069 

8119 

8169 

8219 

8269 

8320 

8370 

8420 

8470 

868 

8520 

8570 

8620 

8670 

8720 

8770 

8820 

8870 

8919 

8970 

869 

9020 

9070 

9120 

9170 

9220 

9270 

9320 

9369 

9419 

9469 

870 

9519 

9569 

9616 

9669 

9719 

9769 

9819 

9869 

9918 

9968 

871 

940018 

0058 

0118 

0168 

0218 

0267 

0317 

0367 

0417 

0467 

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0516 

0566 

0516 

0566 

0716 

0765 

0815 

0865 

0915 

0964 

873 

1014 

1064 

1114 

1163 

1213 

1263 

1313 

1362 

1412 

1462 

874 

1511 

1561 

1611 

1660 

1710 

1760 

1809 

1859 

1909 

1958 

875 

2008 

2058 

2107 

2157 

2207 

2256 

2308 

2355 

2405 

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876 

2504 

2554 

2603 

2633 

2702 

2752 

2801 

2851 

2901 

2950 

877 

3000 

3049 

3099 

3148 

3198 

3247 

3297 

3346 

3396 

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878 

3495 

3544 

3593 

3643 

3692 

3742 

3791 

3841 

3890 

3939 

879 

3989 

4038 

40S8 

4137 

4186 

4236 

4285 

4335 

4384 

4433 

880 

4483 

4532 

4581 

4631 

4680 

4729 

4779 

4828 

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4927 

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4976 

5025 

5074 

5124 

5173 

5222 

5272 

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5370 

5419 

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5469 

5518 

5567 

5616 

5665 

5715 

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5813 

5862 

5912 

833 

5961 

6010 

6059 

6103 

6157 

6207 

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6354 

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6452 

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7041 

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7434 

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7532 

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7728 

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7826 

7875 

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7924 

7973 

8022 

8070 

8119 

8168 

8217 

8266 

8315 

8365 

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8413 

8462 

8511 

8560 

8609 

8657 

8708 

3  755 

8804 

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8902 

8951 

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9048 

9097 

9146 

9195 

9244 

9292 

9341 

890 

9390 

9439 

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9536 

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9634 

9683 

9731 

9780 

9829 

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98  78 

9926 

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..73 

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0351 

0900 

0949 

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1046 

1095 

1143 

1192 

1240 

1239 

894 

1338 

1386 

1435 

1483 

1532 

48. 

2017 

1580 

1629 

1677 

1726 

17/5 

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1823 

1872 

1920 

1969 

2086 

2114 

2163 

2211 

2260 

896 

2303 

2356 

2405 

2453 

2502 

2550 

2599 

2647 

5696 

2744 

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2792 

2841 

2889 

2938 

2986 

3034 

3033 

3131 

3180 

3228 

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3276 

3325 

3373 

3421 

3470 

3518 

3666 

3615 

3663 

3711 

899 

i 

3760 

3803 

3856 

3905 

3953 

4001 

40  i9 

4098 

4146 

4194 

OF  NUMBERS.             19 

N. 

0 

4291 

2 

3 

4 
4435 

5 

6 

7 

8 

9 

900 

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4339 

4387 

4484 

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4725 

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4821 

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5014 

6062 

5110 

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5399 

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5840 

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5688 

5736 

5784 

5832 

5880 

5928 

5976 

6024 

6072 

6120 

904 

6168 

6216 

6265 

6313 

6361 

48 

6840 

6409 

6457 

6505 

6553 

6601 

905 

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6745 

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6888 

6936 

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7032 

7080 

906 

7128 

7176 

7224 

7272 

7320 

7368 

7416 

7464 

7512 

7559 

907 

7607 

7655 

7703 

7751 

7799 

7847 

7894 

7942 

7990 

8038 

903 

8086 

8134 

8181 

8229 

8277 

8325 

8373 

8421 

8468 

8516 

909 

8564 

8612 

8659 

8707 

8755 

8803 

8850 

8898 

8946 

8994 

910 

9041 

9089 

9137 

9185 

9232 

9280 

9328 

9375 

9423 

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911 

9518 

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9614 

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9900 

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912 

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.376 

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913 

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0861 

0709 

0756 

0804 

0851 

0899 

914 

0946 

0994 

1041 

1089 

1136 

1184 

1231 

1279 

1326 

1374 

915 

1421 

1469 

1516 

1563 

1611 

1658 

1706 

1753 

1801 

1848 

916 

1895 

1943 

1990 

2038 

2035 

2132 

2180 

2227 

2275 

2322 

917 

2369 

2417 

2464 

2511 

2559 

2606 

2653 

2701 

2748 

2795 

918 

2843 

2890 

2937 

2985 

3032 

3079 

3126 

3174 

3221 

3268 

919 

3316 

3363 

3410 

3457 

3504 

3552 

3599 

3646 

3693 

3741 

920 

3788 

3835 

3882 

3929 

3977 

4024 

4071 

4118 

4165 

4212 

921 

4260 

4307 

4354 

4401 

4448 

4495 

4542 

4590 

4637 

4684 

922 

4731 

4778 

4825 

4872 

4919 

4966 

5013 

5081 

5108 

5155 

923 

5202 

5249 

5296 

5343 

5390 

5437 

5484 

5531 

5578 

5625 

924 

5672 

5719 

5766 

5813 

5860 

5907 

5954 

6001 

6048 

6095 

925 

6142 

6189 

6236 

6283 

6329 

6376 

6423 

6470 

6517 

6564 

926 

6611 

6658 

6705 

6752 

6799 

6845 

6892 

6939 

6986 

7033 

927 

7080 

7127 

7173 

7220 

7267 

7314 

7361 

7403 

7454 

7501 

928 

7548 

7595 

7642 

7688 

7735 

7782 

7829 

7875 

7922 

7969 

929 

8016 

8062 

8109 

8156 

8203 

8249 

8296 

8343 

8390 

8436 

930 

8483 

8530 

8576 

8823 

8670 

8716 

8763 

8810 

8858 

8903 

931 

8950 

8996 

9043 

9090 

9136 

9183 

9229 

9276 

9323 

9369 

.  932 

9416 

9463 

9509 

9556 

9602 

9649 

9695 

9742 

9789 

9835 

933 

9882 

9928 

9975 

..21 

..63 

.114 

.161 

.207 

.254 

.300 

934 

970347 

0393 

0440 

0486 

0533 

0579 

0626 

0672 

0719 

0765 

935 

0812 

0358 

0904 

0951 

0997 

1044 

1090 

1137 

1183 

1229 

936 

1276 

1322 

1369 

1415 

1461 

1508 

1554 

1601 

1647 

1693 

937 
938 

1740 

1786 

1832 

1879 

1925 

1971 

2018 

2064 

2110 

2157 

2203 

2249 

2295 

2342 

2388 

2434 

2481 

2527 

2573 

2619 

939 

2666 

2712 

2758 

'2804 

2851 

2397 

2943 

2989 

3035 

3082 

940 

3128 

3174 

3220 

3266 

3313 

3359 

3405 

3451 

3497 

3543 

941 

3590 

3836 

3682 

3728 

3774 

3820 

3866 

3913 

3959 

4005 

942 

4051 

4097 

4143 

4189 

4235 

4281 

4327 

4374 

4420 

4466 

943 

4512 

4558 

4604 

4650 

4898 

4742 

4788 

4834 

4380 

4926 

944 

4972 

5018 

5084 

5110 

5156 

46 

5616 

5202 

5248 

5294 

5340 

5386 

945 

5432 

5478 

5524 

5570 

5662 

5707 

5753 

5799 

5845 

946 

5891 

5937 

5983 

6029 

6075 

6121 

6167 

6212 

6258 

6304 

947 

6350 

6396 

6442 

6488 

6533 

6579 

6925 

6671 

6717 

6763 

948 

6803 

6854 

6900 

6946 

6992 

7037 

7033 

7129 

7175 

7220 

949 

7266 

7312 

7358 

7403 

7449 

7495 

7541 

7586 

7632 

7678 

19 


20 

LOGARITHMS 

N. 

0 

1 

2 

3 

4 

5 

6 

7 
8043 

8 

9 

950 

977724 

7769 

7815 

7861 

7906 

7952 

7998 

8089 

8135 

951 

8181 

8226 

8272 

8317 

£363 

8409 

8454 

8500 

8546 

8591 

952 

8637 

8683 

8728 

8774 

8819 

8865 

8911 

8956 

9002 

9047 

953 

9093 

9138 

9184 

9230 

9275 

9321 

9366 

9412 

9457 

9503 

954 

9548 

9594 

9639 

9685 

9730 

46 

0185 

9776 

9821 

9867 

9912 

9968 

955 

980003 

0049 

0094 

0140 

0231 

0276 

0322 

0367 

0412 

956 

0458 

0503 

0549 

0594 

0340 

0685 

0730 

0776 

0321 

0867 

957 

0912 

0957 

1003 

1048 

1093 

1139 

1184 

1229 

1275 

1320 

958 

1366 

1411 

1456 

1501 

1547 

1592 

1637 

1683 

1728 

1773 

959 

1819 

1864 

1909 

1954 

2000 

2045 

2090 

2135 

2181 

2226 

960 

2271 

2316 

2362 

2407 

2452 

2497 

2543 

2588 

2633 

2678 

961 

2723 

2769 

2814 

2859 

2904 

2949 

2994 

3040 

3085 

3130 

962 

3175 

3220 

3265 

3310 

3356 

3401 

3446 

3491 

3536 

3581 

963 

3626 

3671 

3716 

3762 

3807 

3852 

3897 

3942 

3987 

4032 

964 

4077 

4122 

4167 

4212 

4257 

43^2 

4347 

4392 

4437 

4482 

965 

4527 

4572 

4617 

4662 

4707 

4752 

4797 

4842 

4887 

4932 

966 

4977 

5022 

5087 

5112 

5157 

5202 

5247 

5292 

5337 

5382 

96; 

5426 

5471 

5516 

5561 

5606 

5651 

5699 

5741 

5786 

5830 

968 

5875 

5920 

5965 

6010 

6055 

6100 

6144 

6189 

6234 

6279 

969 

6324 

6369 

6413 

6458 

6503 

6548 

6593 

6637 

6682 

6727 

970 

6772 

6817 

6861 

6908 

6951 

6996 

7040 

7035 

7130 

7175 

971 

7219 

7264 

7309 

7353 

7398 

7443 

7488 

7532 

7577 

7622 

972 

7666 

7711 

7756 

7800 

7845 

7890 

7934 

7979 

8024 

8068 

973 

8113 

8157 

8202 

8247 

8291 

8336 

8381 

8425 

8470 

8514 

974 

8559 

8604 

8648 

8693 

8737 

8782 

8826 

8871 

8916 

8960 

975 

9005 

9049 

9093 

9138 

9183 

9227 

9272 

9316 

9361 

9405 

976 

9450 

9494 

9539 

9583 

9628 

9672 

9717 

9761 

9806 

9850 

977 

9895 

9939 

9983 

..28 

..72 

.117 

.161 

.206 

.250 

.294 

978 

990339 

0383 

0428 

0472 

0516 

0561 

0605 

0650 

0694 

0738 

979 

0783 

0827 

0871 

0916 

0960 

1004 

1049 

1093 

1137 

1182 

980 

1226 

1270 

1315 

1359 

1403 

1448 

1492 

1536 

1580 

1625 

981 

1669 

1713 

1758 

1802 

1846 

1890 

1935 

1979 

2023 

2067 

982 

2111 

2156 

2200 

2244 

2288 

2333 

2377 

2421 

2465 

2509 

983 

2554 

2598 

2642 

2686 

2730 

2774 

2819 

2863 

2907 

2951 

984 

2995 

3039 

3083 

3127 

3172 

3216 

3260 

3304 

3348 

3392 

985 

3436 

3480 

3524 

3568 

3613 

3657 

3701 

3745 

3789 

3833 

986 

3877 

3921 

3965 

4009 

4053 

4097 

4141 

4185 

4229 

4273 

987 

4317 

4361 

4405 

4449 

4493 

4537 

4581 

4625 

4669 

4713 

988 

4757 

4801 

4845 

4886 

4933 

4977 

5021 

5065 

5108 

5152 

989 

5196 

5240 

5284 

5328 

5372 

5416 

5460 

5504 

5547 

5591 

:  990 

5635 

5379 

5723 

5767 

5811 

5854 

5898 

5942 

5986 

6030 

:  991 

6074 

6117 

6161 

6205 

6249 

6293 

6337 

6380 

6424 

6468 

I  992 

6512 

6555 

6599 

6643 

6687 

6731 

6774 

6818 

6862 

6906 

993 

6949 

6993 

7037 

7080 

7124 

7168 

7212 

7255 

7299 

7343 

994 

7386 

7430 

7474 

7517 

7561 

44 

7998 

7605 

7648 

7692 

7736 

7779 

995 

7823 

7867 

7910 

7954 

8041 

8085 

8129 

8172 

8216 

996 

8259 

8303 

8347 

8390 

8434 

8477 

8521 

8564 

8608 

8652 

997 

8695 

8739 

8792 

8826 

8869 

8913 

8956 

9000 

9043 

9087 

998 

9131 

9174 

9218 

9261 

9305 

9348 

9392 

9435 

9479 

9522 

999 

9565 

9609 

9652 

9698 

9739 

9783  |  9826 

9870 

9913 

9957 

r -=? 

TABLE  If.    Log.  Sines  and  Tangems.  (0°)  Natural  Sines          21 

0 

Sine.   D  10" 

Cosine. 

D.IO" 

Tang. 

D.IO" 

Coiang. 

N.sine 

N.  cos. 

9.OJ0000 

10.000000 

0.000000 

Infinite. 

00001 

100000 

60 

1 

6. 463726 

000000 

6.463726 

13.536274 

0002'd 

100000 

59 

2 

764766 

039000 

764756 

235244 

00058 

iooooo 

58 

3 

940847 

009000 

940847 

059153 

00087 

100000 

57 

4 

7.065786 

000)0  J 

7.065786 

12.934214 

00116 

100000 

56 

5 

162696 

000003 

162696 

837304 

00145 

100000 

55 

6 

241877 

9.999999 

241878 

758122 

00175 

100000 

54 

7 

308824 

999999 

308825 

691175 

00204 

100000 

53 

366816 

999999 

366817 

633183 

00233 

100000 

52 

9 

417968 

999999 

417970 

582030 

00262 

100000 

51 

10 

463725 

999998 

463727 

536273 

00291 

100000 

50 

11 

7.505118 

9.999998 

7.505120 

12.494880 

00320 

99999 

49 

1-2 

542905 

999997 

542909 

457091 

00349 

99999 

48 

13 

577668 

999997 

577672 

422328 

00378 

99999 

47 

14 

609853 

999996 

609857 

390143 

00407 

99999 

46 

15 

639816 

999996 

639820 

360180 

00436 

99999 

45 

16 

667845 

999995 

667849 

332151 

00465 

99999 

44 

17 

694173 

999995 

694179 

305821 

00495 

99999 

43 

18 

718997 

999994 

719003 

280997 

00524 

99999 

42 

19 

742477 

999993 

742484 

257516 

00553 

99998 

41 

20 

764754 

999993 

764761 

235239 

00582 

99998 

40 

21 

7.785943 

9.999992 

7.785951 

12.214049 

00611 

99998 

39 

22 

806146 

999991 

808155 

193845 

00640 

99998 

38 

23 

825451 

999990 

825460 

174540 

00669 

99998 

37 

24 

843934 

999989 

843944 

156056 

00698 

99998 

36 

25 

861663 

999988 

861674 

138326 

00727 

99997 

35 

26 

878695 

999988 

878708 

121292 

00756 

99997 

34 

27 

895085 

999987 

895099 

104901 

00785 

99997 

33 

28 

910879 

999986 

910894 

089106 

00814 

99997 

32 

29 

926119 

999985 

926134 

073866 

00844 

99996 

31 

30 

940842 

999983 

940858 

059142 

00873 

99996 

30 

31 

7.955082 

2298 
2227 
2161 
2098 
2039 
1983 
1930 
1880 
1832 
1787 
1744 
1703 
1664 
1626 
1591 
1557 
1524 
1492 
1462 
1433 
1405 
1379 
1353 
1328 
1304 
1281 
1259 
1237 
1216 

9.999982 

7.955100 

2298 
2227 
2161 
2098 
2039 
1983 
1930 
1880 
1833 
1787 
1744 
1703 
1664 
1627 
1591 
1557 
1524 
1493 
1463 
1434 
1406 
1379 
1353 
1328 
1304 
1281 
1259 
1238 
1217 

12 . 044900 

00902 

99996 

29 

32 

968870 

999981 

0.2 
0.2 
0.2 

968889 

031111 

00931 

99996 

28 

,33 

982233 

999980 

982253 

017747 

00960 

99995" 

27 

34 

995198 

999979 

995219 

004781 

00989 

99995 

26 

35 

8.007787 

999977 

0-2 
0-2 
0-2 

8.007809 

11.992191 

01018 

99995 

25 

36 

020021 

999976 

020045 

979955 

01047 

99995 

24 

37 

031919 

999975 

031945 

968055 

01076 

99994 

23 

38 

043501 

999973 

02 
0-2 

043527 

956473 

01105 

99994 

22 

39 

054781 

999972 

054809 

945191 

01134 

99994 

21 

40 

065776 

999971 

0'2 
0'2 

065806 

934194 

01164 

99993 

20 

41 

8.076500 

9.999969 

8.076531 

11.923469 

01193 

99993 

19 

42 

086965 

999968 

0  2 
0*2 
0'2 
0'3 
0'3 

o's 

0.3 
0.3 
0.3 
0.3 
0.3 
0.3 
0.3 
0.3 
0.3 
0.4 
0.4 
0.4 
0.4 

086997 

913003 

01222 

99993 

18 

43 

097183 

999966 

097217 

902783 

01251 

99992 

17 

44 

107167 

999964 

107202 

892797 

01280 

99992 

16 

45 

116926 

999963 

116963 

883037 

01309 

99991 

15 

46 

126471 

999961 

126510 

873490 

01338 

99991 

14 

47 

135810 

999959 

135851 

864149 

01367 

99991 

13 

48 

144953 

999958 

144996 

855004 

01396 

99990 

12 

49 

153907 

999956 

153952 

846048 

01425 

99990 

11 

50 

162681 

999954 

162727 

837273 

01454 

99989 

10 

51 

8.171280 

9.999952 

3.171328 

11.828672 

01483 

99989 

9 

52 

179713 

999950 

179763 

820237 

01513 

99989 

8 

53 

187985 

999948 

188036 

811964 

01542 

99988 

7 

54 

196102 

999946 

196156 

803844 

01571 

99988 

6 

55 

204070 

999944 

204126 

795874  101600 

99987 

5 

56 

211895 

999942 

211953 

788047  j 01629 

99987 

4 

57 
58 

219581 
227134 

999940 
999938 

219641 
227195 

780359  101658 
772805  |i  01687 

99986 
99986 

3 

2 

59 

234557 

999936 

234621 

765379  !|  01716 

99985 

1 

60 

241855 

999934 

241921 

758079 j 

01745 

N.  cos. 

99985 
N.  sine- 

0 

~ 7 

Cosine. 

Sine. 

Cot  an?. 

Tan  jr. 

89  Degrees. 

;22 


Lo-r.  Sines  and  Ta 


0°) 


TABLE 


|D  10' 


0 
1 

2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
1? 
18 
19 
20 
21 
22' 
23 
24 
25 
26 
2? 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 

as 

39 
40 
41 
42 
43 
44 
45 
40 
47 
48 
49 
59 
51 
52 
53 
54 
55 
53 
57 
58 
59 
60 


3.241855  11Qp 
249033  }{^2 
256094  {}'' 
263042  }}» 
269881  [{$" 
276614  \\jt 
283243  }££ 
*°3' ia   1079 

296207  Jx-fi 

302546  ^ 

308794  Jnoi 

$.314954  {xf* 

321027  qQ« 

327016  gg° 

332924  g*° 

338753  q-q 

344504  HI 

350181  ™ 

355783  qS 

361315  qTq 

366777  goo 

5.372171  o«« 

377499  o„ 

382762  2' ' 

387962  o~' 

393101  g™ 

398179  007 

4031991  qJ7 

408161 !  o7o 

4130381  o,,t; 

417919  jJJn 

$.422717  °g° 

427462  ^ 

432156  ,~ 

436800!  «^ 

UlSt  I  7^ 

4o0440  ?42 

454893  '™ 

459301  !  Lfrj 

463665  i^n 

$.467985  ifi 

472263  !  ^ 

476498  '$ 

480693  *£ 

484848  ££ 

488963  ^°q 

493040  ^ 

497078  ^ 

501080  }*' 

505045  w» 

$.508974  ™? 

512867  £** 

516726  °~ 

520551  ™7 

524343  ^ 

528102  Jo? 

531828  °f; 

535523  °|° 

539186  J?.1* 

542819  UD 


Cosine. 


D.10' 


0.4 
0.4 
0.4 
0.4 
0.4 
0.4 
0.4 
0.4 
0.4 
0-4 
0-4 


0-5 
05 
0-5 
0.5 
0.5 
0-5 
05 
0.5 
0.5 
0-5 
0.5 
0.5 
0.5 
0-5 
06 
0-6 
0.6 
0-6 
0-6 
0.6 
0.6 
0.6 
0-6 
0-6 
0-6 
0-6 
0-6 
0.6 
0.6 
0-7 
07 


0.7 
0.7 
0.7 
0.7 
0.7 
0.7 


$.241921 
249102 
256165 
263115 
269956 
276691 
283323 
289856 
298292 
302634 
308884 

$.315046 
321122 
327114 
333025 
333856 
344610 
35U289 
355895 
361430 
366895 

t.  372292 
377622 
382889 
388092 
393234 
398315 
403338 
408304 
413213 
418088 

;.  422869 
427618 
432315 
436962 
441560 
446110 
450613 
455070 
459481 
463849 

.468172 
472454 
476693 
480892 
485050 
489170 
493250 
497293 
501298 
505267 

.509200 
513098 
516981 
520790 

. 524586 
528349 
532089 
535779 
539447 
543084 

"Couing-. 


D  10' 


1197 
1177 
1158 
1140 
1122 
1105 
1089 
1073 
1057 
1042 
1027 
1013 
999 
985 
972 
959 
946 
934 
922 
911 
899 
888 
879 
867 
857 
847 
837 
828 
818 
809 
800 
791 
783 
774 
766 
758 
750 
743 
735 
728 
720 
713 
707 
700 
693 
686 
680 
674 
668 
661 
655 
650 
644 
638 
633 
627 
622 
616 
611 
606 


G 


'N.  sine.  X.  cos.l 


11 


11 


11.758079 
750898 
743835 
736885 
730044 
723309 
716677 
710144 
703708 
697366 
691116 
684954 
678878 
672886 
666975 
661144 
655390 
649711 
644105 
638570 
633105 
627708 
622378 
617111 
611908 
606766 
601685 
596662 
591696 
586787 
581932 

11.577131 
572382 
567685 
563038 
558440 
553890 
549387 
544930 
540519 
536151 

11-531828 
527546 
523307 
519108 
514950 
510830 
506750 
502707 
498702 
494733 

11.490800 
486902 
483039 
479210 
475414 
471651 
467920 
464221 
460553 ; 
456916 , 
Tana:.   I 


0174 
01774 
0180; 
01832 
01862 
01891 
01920 
0194! 
01978 
02007 
02036 
02065 
02094 
02123 
02152 
02181 
02211 
02240 
02269 
02298 
0232 
02356 
02385 
02414 
02443 
02472 
02501 
02530 
02560 
02589 
02618 
0264; 
02676 
02705 
02734 
02763 
02792 
02821 
02850 
02879 
02908 
02938 
02967 
02996 
03025 
03054 
03083 
03112 
03141 
03170 
03199 
03228 
03257 
! 03286 
'<  03316 
■  03345 
: 03374 
i  03403 
: 03432 
03461 
, 03490 


99385;  60 
)9984i  59 
99934  58 
99983!  57 
999831  56 
99982  55 
99982  54 
99^81  63 
99980  52 
99980  51 
99979'  50 
Jy979j  48 
y9978|  48 
99977  47 
);  977  46 
99976  45 
99976  44 
^9975  43 
99974^  42 
99974!  41 
99973  40 
b972  39 
99972  38 
99971  37 
99970  36 
89969  35 
99969  34 
99968'  33 
99967|  32 
99966!  31 


;9966 
99985 
99984 
99963 
99963 
99962 
99961 
99960 
99959 
99959 
99958 
99957 
99956 
99955 
99954 
99953 
99952 
99952 
99951 
99950 
99949 
99948 
99947 
99946 
99945 
99944 
99943 
99942 
99941 
99940 
999b9 


X.  cos.  X.sine.  ' 


83  Degrees. 


TARLK  II.    Log.  Sines  and  Tangents.  (-2°)  Natural  Sines. 


23 


0 
1 
2 
3 
4 
5 
6 
7 
S 
9 
10 

11 

12 

13 
14 
15 
16 

17 

18 

IS 
20 

•21 
2^ 
23 
21 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
GO 


8.542819 
546422 
549995 
553539 
557054 
560540 
563999 
567431 
570836 
574214 
577566 

8.580892 
584193 
587469 
590721 
593948 
597152 
600332 
603489 
606623 
609734 

8.612823 
615891 
618937 
621962 
624965 
627948 
630911 
633854 
636776 
639680 

8.642563 
645428 
648274 
651102 
653911 
656702 
659475 
662230 
664968 
667689 

8.670393 
673080 
675751 
678405 
681043 
683665 
686272 
688863 
691438 
693998 

8.696543 
699073 
701589 
704090 
706577 
709049 
711507 
713952 
716383 
718800 
Cosine. 


D.  10 

600 
595 
591 
586 
581 
576 
572 
567 
563 
559 
554 
550 
546 
542 
538 
534 
530 
526 
522 
519 
515 
511 
508 
504 
501 
497 
494 
490 
487 
484 
481 
477 
474 
471 
468 
465 
462 
459 
456 
453 
451 
448 
445 
442 
440 
437 
434 
432 
429 
427 
424 
422 
419 
417 
414 
412 
410 
407 
405 
403 


9.999735 
999731 
999726 
999722 
999717 
999713 
999708 
999704 
999899 
999694 
999689 

9.999685 
999680 
999675 
999670 
999665 
999660 
999655 
999650 
999645 
999640 
999635 
999629 
999324 
999619 
999614 
999608 
999603 
999597 
999592 
999586 
^.999581 
999575 
999570 
999564 
999558 
999553 
999547 
999541 
999535 
999529 
999524 
999518 
999512 
999506 
999500 
999493 
999487 
999481 
999475 
999469 
999463 
939456 
999450 
999443 
999437 
999431 
999424 
999418 
999411 
999404 


D.  10y 


0.7 
0.7 
0-7 

0-8 
0-8 
0-8 
0.8 
0.8 
0-8 
0-8 
0.8 
0-8 
0-8 
0.8 
0.8 
0.8 
0.8 
0.8 
0.8 
0.8 
0.9 


Tansr. 


0.9 
0.9 
0-9 
0-9 
0-9 
0-9 
0-9 
0-9 
0.9 
0-9 
0.9 
0.9 
0.9 
0.9 


1 

1 

1 

1-0 

1.0 

1-0 

1-0 

1.0 

1.0 

1.0 

1.0 

1.0 

1.0 

1.1 

1 
1 
1 
1 
1 
1 


8.543084 
546691 
550268 
653817 
557336 
560828 
564291 
567727 
571137 
574520 
577877 

8.581208 
584514 
587795 
591051 
594283 
597492 
600677 
603839 
608978 
610094 

8.613189 
616262 
619313 
622343 
625352 
628340 
631308 
634256 
637184 
640093 

8.642982 
645853 
648704 
651537 
654352 
657149 
659928 
662689 
665433 
668160 

8.670870 
673563 
676239 
678900 
681544 
684172 
6  6784 
689381 
691963 
694529 
.697081 
699617 
702139 
704246 
707140 
709618 
702083 
714534 
716972 
719396 
"Coiaiisr. 


D.  W 

602 

59;} 

591 

587 

582 

577 

573 

568 

564 

559 

555 

551 

547 

543 

539 

535 

531 

527 

523 

519 

516 

512 

508 

505 

501 

498 

495 

491 

488 

485 

482 

478 

475 

472 

469 

466 

463 

460 

457 

454 

453 

449 

446 

443 

442 

438 

435 

433 

430 

428 

425 

423 

420 

418 

415 

413 

411 

408 

406 

404 


Cotang.  |  IN.  sine 


11 


11.456916 
453309 
449732 
446183 
442664 
439172 
435709 
432273 
428863 
425480 
422123 
,418792 
415486 
412205 
408949 
405717 
402508 
399323 
396161 
393022 
389906 
11.386811 
383738 
380687 , 
377657  I 
374648 
371660 
368692 
365744 
362816 
359907 
11.357018 
354147 
351296 
348463 ! 
345648  j 
342851 i 
340072 
337311 
334567 
331840 
11.329130 
326437 
323761 
321100 
318456 
315828 
313216 
310819 
308037 
305471 
11.302919 
300383  j 
297861 j 
295354 | 
292860 ; 
290382  i 
287917  ! 
285465 ! 
283028 
280604 


03490 
03519 
03548 
03577 
03606 
03635 
,03664 
03693 
03723 
03752 
03781 
03810 
03839 
03868 
03897 
03926 
03955 
03984 
04013 
04042 
04071 
04100 
03129 
04159 
04188 
04217 
04246 
04275 
04304 
04333 


N.  cos 


99.939 
99938 
99937 
99936 
99935 
99934 
99933 
99932 
99931 
99930 
99929 
99927 
99926 
99925 
99924 
99923 
99922 
99921 
99919 
99918 
99917 
99916 
99915 
99913 
99912 
99911 
99910 
99909 
99907 
99908 


04362J99905 


04391 
04420 
04449 
04478 
04507 
04536 


04565-99896 


04594 
04623 
04653 
04682 
04711 
04740 
04769 
04798 
04827 
04856 
04885 
04914 
04943 
04972 
05001 
05030 
05059 
05088 
05117 
05146 
05175 
05205 
05234 


N.  cos.  N.siire 


99904 
99902 
99901 
99900 
99898 
9989' 


99894 
99893 
99892 
99890 
99889 
99888 
99886 
99885 
99883 
99882 
99881 
99879 
99878 
99876 
99875 
99873 
99872 
99870 
99869 
99867 
99866 
99864 
99863 


60 
59 

58 
57 
56 
55 
54 
58 
52 
51 
5C 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
35 
34 
33 
32 
31 
80 
29 
28 
27 
28 
25 
24 
23 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 

10 
9 

8 
7 
6 
5 

4 
3 
2 

1 
0 


87  Degrees. 


24 


Log.  Sines  and  Tangents.     (3°;     Natural  Sines.        TABLE  II. 


0 
1 

2 
3 
4 
5 
6 
7 
8 
B 
10 

11 

12 

13 
14 
15 
16 
17 
18 
& 
20 
21 
22 
23 
24 
25 
26 
27 
38 
29 
30 
31 
32 
33 
34 
35 
30 
3? 
38 
39 
40 
41 
42 
43 
44 
45 
46 
4? 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 


Shit-. 

i.  718809 
721204 
723595 
725972 
728337 
730888 
733027 
735354 
737667 
739969 
742259 

1.744536 
746802 
749055 
751297 
753528 
755747 
757955 
760151 
762337 
764511 

;.  766675 
768828 
770970 
773101 
775223 
777333 
779434 
781524 
783605 
785675 

1.787736 
789787 
791828 
793859 
795881 
797894 
799897 
801892 
803876 
805852 

i.  8078 19 
809777 
811726 
813667 
815599 
817522 
819436 
821343 
823240 
825130 

1.827011 
828884 
830749 
832607 
834456 
836297 
838130 
839956 
841774 
843585 


Cosine. 


401 
398 
396 
394 
392 
390 
388 
386 
384 
382 
380 
378 
376 
374 
372 
370 
368 
366 
364 
362 
361 
359 
357 
355 
353 
352 
350 
348 
347 
345 
343 
342 
340 
339 
337 
335 
334 
332 
331 
329 
328 
326 
325 
323 
322 
320 
319 
318 
316 
315 
313 
312 
311 
309 
308 
307 
306 
304 
303 
302 


Cosine. 

.999404 
999398 
999391 
999384 
999378 
999371 
999364 
999357 
999350 
999343 
999336 

.999329 
999322 
999315 
999308 
999301 
999294 
999286 
999279 
999272 
999265 

.999257 
999250 
999242 
999235 
999227 
999220 
999212 
999205 
999197 
999189 

.999181 
999174 
999166 
999158 
999150 
999142 
999134 
999126 
999118 
999110 

.999102 
999094 
999086 
999077 
999069 
999061 
999053 
999044 
999036 
999027 

.999019 
999010 
999002 
998993 
998984 
998976 
998967 
998958 
998950 
998941 


Sin«\ 


D.  10'       Cotang.  |(N.sine 


.719396 
721806 
724204 
726588 
728959 
731317 
733663 
735996 
738317 
740S26 
742922 

.745207 
747479 
749740 
751989 
754227 
756453 
758668 
760872 
763065 
765246 

.767417 
769578 
771727 
773866 
775995 
778114 
780222 
782320 
784408 
786486 

.788554 
790613 
792662 
794701 
796731 
798752 
800763 
802765 
804858 
806742 

.808717 
810683 
812641 
814589 
816529 
818461 
820384 
822298 
824205 
826103 

.827992 
829874 
831748 
833613 
835471 
837321 
839163 
840998 
842825 
844644 

Cotang. 


402 
399 
397 
395 
393 
391 
389 
387 
385 
383 
381 
379 
377 
375 
373 
371 
369 
367 
365 
364 
362 
360 
358 
356 
355 
353 
351 
350 
348 
346 
345 
343 
341 
340 
338 
337 
335 
334 
332 
331 
329 
328 
326 
325 
323 
322 
320 
319 
318 
316 
315 
314 
312 
311 
310 
308 
307 
306 
304 
303 


11.280604 
278194 
275796 
273412 
271041 
268683 
266337 
264004 
261683 
259374 
257078 

11.254793 
252521 
250260 
248011 
245773 
243547 
241332 
239128 
236935 
234754 

11.232583 
230422 
228273 
226134 
224005 
221886 
219778 
217680 
215592 
213514 

11.211446 
209387 
207338 
205299 
203269 
201248 
199237 
197235 
195242 
193258 
191283 
189317 
187359 
185411 
183471 
181539 
179616 
177702 
175795 
173897 

11.172008 
170126 
168252 
166387 
164529 
162679 
160837 
159002 
157175 
155356 


05234 
05263 
05292 
05321 
05350 
05379 
05408 
05437 
05466 
05495 
05524 
05553 
05582 
05611 
05640 
05669 
05698 
05727 
05756 
05785 
05814 
05844 
05873 
05902 
05931 
05960 
05989 
06018 
06047 
06076 
06105 
06134 
06163 
06192 
08221 
06250 
06279 
06308 
06337 
06366 
06395 
08424 
06453 
06482 
06511 
08540 
08569 
06598 
0662' 
,06656 
06685 
06714 
06743 
06773 
06802 
1 06831 
1 06860 
! 06889 
1 06918 
|  0694 
!  06976 


Tang.   II  N.  cos.  X.sine. 


N.  cos 


99863 
99861 
99860 
99858 
99857 
99855 
99854 
99852 
99851 
99849 
99847 
99846 
99844 
99842 
99841 
99839 
99838 
99836 
99834 
99833 
99831 
99829 
99827 
99826 
99824 
99822 
99821 
99819 
99817 
99815 
99813 
99812 
99810 
99808 
99806 
99804 
99803 
99801 
99799 
99797 
99795 
99793 
99792 
99790 
99788 
99786 
99784 
99782 
99780 
99778 
99776 
99774 
99772 
99770 
99768 
99766 
99764 
99762 
99760 
99758 
99756 


86  Degrees. 


TABLE  H.        Log.  Sines  and  Tangents.     (4°)     Naiu.-al  Sines. 


Sino. 


8.843585 
845387 
847183 
848971 
850751 
852525 
854291 
856049 
857801 
859546 
861283 

8.863014 
864738 
866455 
868165 
869868 
871565 
873255 
874938 
876615 
878285 

8.879949 
881607 
883258 
884903 
886542 
888174 
889801 
891421 
893035 
894643 

8.896246 
897842 
899432 
901017 
902596 
904169 
905736 
907297 
908853 
910404 

8.911949 
913488 
915022 
916550 
918073 
919591 
921103 
922610 
924112 
925609 

8.927100 
928587 
930088 
931544 
933015 
934481 
935942 
937398 
938850 
94u296 
Cosine 


1).  10" 


287 
286 
285 
284 
283 
282 
281 
279 
279 
277 
276 
275 
274 
273 
272 
271 
270 
269 
268 
267 
266 
265 
264 
263 
262 
261 
260 
259 
258 
257 
257 
256 
255 
254 
253 
252 
251 
250 
249 
249 
248 
247 
248 
245 
244 
243 
243 
242 
241 


Cosine. 

1.998941 
998932 
998923 
998914 
998905 
998896 
998887 
998878 
998869 
998860 
993851 

1.998841 
998832 
998823 
998813 
993804 
998795 
998785 
998776 
998766 
998757 

1.998747 
998738 
998728 
998718 
998708 
998699 
998689 
998679 
998669 
998659 

1.998649 
998639 
998629 
998619 
998609 
998599 
998589 
998578 
998568 
998558 

.998548 
998537 
998527 
998516 
998508 
998495 
998485 
998474 
998464 
998453 

.998442 
998431 
998421 
998410 
998399 
f 98388 
998377 
998366 
998355 
998344 


Sine. 


I).  10" 


1.5 
1.5 
1.5 
1.5 
1-5 
1-5 
1.5 
1.5 
1-5 
1-5 
1.5 
1.5 
1-5 
1.6 


1 
1 

1 

1 
1 

1 
1 

J 

1 

1 

1.6 

1-6 

1.6 

1.7 

1.7 

1.7 

1.7 

1.7 

1.7 

1.7 

1.7 

1.7 

1.7 

17 

1.7 

1.7 

1.7 

1.7 

1.8 


1.8 

1.8 

1.8 

1.8 

1.8 

1 

1 


8 

8 

1.8 


Tang. 


>. 844644 
846455 
848260 
850057 
851846 
853628 
855403 
857171 
858932 
860686 
862433 

5.864173 
865906 
867 632 
869351 
871064 
872770 
874469 
876162 
877849 
879529 

i. 881202 
882869 
884530 
886185 
887833 
889476 
891112 
892742 
894366 
895984 

i. 897596 
899203 
900803 
902398 
903987 
905570 
907147 
908719 
910285 
911846 

i. 91 3401 
914951 
916495 
918034 
919568 
921096 
922619 
924136 
925649 
927156 

!. 928658 
930155 
931647 
933134 
934616 
936093 
937565 
939032 
940494 
941952 

Cotanjr. 


D.  10/ 

302 
301 
299 
298 
29/ 
29  > 
295 
293 
292 
291 
290 
289 
288 
287 
285 
284 
283 
282 
281 
280 
279 
278 
277 
276 
275 
274 
273 
272 
271 
270 
269 
268 
267 
266 
265 
264 
263 
262 
261 
260 
259 
258 
257 
256 
256 
255 
254 
253 
252 
251 
250 
249 
24y 
248 
247 
246 
245 
244 
244 
243 


Cotang.  :  N.  sine.  X.  cos 


11.155356 
153545 
151740 
149943 
148164 
146372 
144597 
142829 
141068 
139314 
137567 

11.135827 
134094 
132368 
130649 
128936 
127230 
125531 
123838 
122151 
120471 

11.118798 
117131 
115470 
113815 
112167 
110524 
108888 
1 07258 
105634 
104016 

11.102404 
100797 
099197 
097602 
096013 
094430 
092853 
091281 
089715 
088154 

11.086599 
085049 
083505 
081966 
080432 
078904 
077381 
075864 
074351 
072844 

11.071342 
089845 
068353 
066866 
065384 
063907 
062435 
060968 
059506 
058048 
Tang-. 


06976 
07005 
07034 
07083 
07092 
07121 
07150 
07179 
07208 
07237 
07266 


99756 

99754 

99752 

99750 

99748 

99746 

99744 

99742 

99740 

99738! 51 

99736  50 


07295  99734 


i1  08339  99652 
•108368  99649 

I  08397  99647 
'i  08426  99644 
;  08455  99642 
|:  08484  99689 
•0851399637 

1 08542  99635 
:'  0857 199632 
1 108600  99630 

II  08629  9y627 
1 108658  99625 

08687  99622 
108716  99619 


49 


07324  99731 

07353  99729 

07382  99727 

0741199725 

07440  99723 

07469  99721 

07498  99719 

07527  99716 

07556  99714 

07585  99712 

07614  99710 

07643  99708 

j  07672  99705 

!  107701  99703 

! i  07730  99701J  34 

! ;  07759  99699  33 

1 107788  99696:82 

!i  07817  99694  31 

1 1  07846  99692  30 

!!  07875  996891  29 

i1  07904  99387,  28 

1 107933  99885  27 

;  1 07962  99683 

10799199680 

!  108020  99878 

,,08049  99676 

i!  08078  99873 

J  08107  4*9671 

108136  99668 

1108165  99666!  19 

!  |  08194  99864'  18 

;i  08223  99661 1  17 

j  J08252  99659  16 

!;  08281  99657|  15 

108310  99654  14 


.\.  cus.  N.sine.  ' 


85.  Degrees. 


26 


Log.  Sines  and  Tangents.     (5°)     Natural   Sines. 


—11 

tabu:  i?. 


D.  10" 


3.940290 
941 738 

943174 
944693 
946034 
947453 
948874 
950287 
951693 
953100 
954499 

3.955894 
957284 
958370 
960052 
961429 
962801 
964170 
965534 
968893 
968249 

3.969800 
970947 
972289 
973628 
974932 
976293 
977619 
978941 
980259 
981573 

8.982S33 
984189 
985491 
988789 
938083 
989374 
990360 
991943 
993222 
994497 

3.995768 
997036 
998299 
999580 

3.009816 
002039 
003318 
004563 
005805 
097044 

9.003278 
009510 
010737 
011982 
013182 
014400 
015613 
016824 
018031 
019235 
Cosine. 


240 
239 
239 
238 
237 
236 
235 
235 
234 
233 
232 
232 
231 
230 
229 
229 
228 
227 
227 
226 
225 
224 
224 
223 
222 
222 
221 
220 
220 
219 
218 
218 
217 
216 
216 
215 
214 
214 
213 
212 
212 
211 
211 
210 
209 
209 
203 
208 
207 
203 
203 
205 
205 
204 
203 
203 
2U2 
202 
201 
201 


9. 


993344 
993333 
998322 
993311 
998300 
998289 
998277 
998266 
998255 
998243 
998232 
.998220 
998209 
998197 
998186 
998174 
998163 
998151 
998139 
998128 
998116 
.998104 
998092 
998030 
998088 
998053 
998044 
998032 
998020 
998008 
997998 
.997984 
997972 
997959 
997947 
997935 
997922 
997910 
997897 
997885 
997872 
.997860 
997847 
997835 
997822 
997809 
997  797 
997784 
997771 
997758 
997745 
.997732 
997719 
997706 
997693 
997680 
997667 
997654 
997641 
997628 
997614 
Sine. 


D.  10' 


1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
1.9 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.0 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.1 
2.2 
2.2 
2.2 
2.2 
2.2 
2.2 


.941952 
943404 
944852 
948295 
947734 
949168 
950597 
952021 
953441 
954856 
956267 

.957674 
959075 
960473 
981866 
983255 
984639 
966019 
967394 
968766 
970133 

.971496 
972855 
974209 
975560 
976908 
978248 
979586 
980921 
982251 
983577 

.984899 
986217 
987532 
988842 
990149 
991451 
992750 
994045 
995337 
996624 

.997908 
999188 

.000465 
001738 
003007 
004272 
005534 
008792 
008047 
009298 

•010546 
011790 
013031 
014268 
015502 
016732 
017959 
019183 
020403 
021620 


Cotang. 


0.  10"|     Coiang.       N.  sine. 


242 
241 
240 
240 
239 
238 
237 
237 
236 
235 
234 
234 
233 
232 
231 
231 
230 
229 
229 
228 
227 
226 
22S 
225 
224 
224 
223 
222 
222 
221 
220 
220 
219 
218 
218 
217 
216 
216 
215 
215 
214 
213 
213 
212 
211 
211 
210 
210 
209 
208 
208 
207 
207 
206 
206 
205 
204 
204 
203 
203 


ill.  058048 
05i>596 
055148 
053705 
052266 
050832 
049403 
047979 
046559 
045144 
043733 

11.042326 
040925 
039527 
038134 
036745 
035361 
033981 
032606 
031234 
029867 

11.028504 
027145 
025791 
024440 
023094 
021752 
020414 
019079 
017749 
016423 

11.015101 
013783 
012468 
011158 
009851 
008549 
007250 
005955 
004663 
003376 

11.002092 
000812 

t0. 999535 
998262 
996998 
995728 
994466 
993208 
991953 
990702 

10.989454 
988210 
686969 
985732 
984498 
983268 
983041 
980817 
979597 
978380 


0871699619 

08745  99617 


i 087 74 
108808 
! 08831 
108860 


99614 
99612 

99609 
99607 


0888999604 


:  108918 
jl  08947 
!  J  08976 
09006 
! !  09034 
1 1 09063 
;  109092 
'109121 


99802 
99599 
99596 
99594 
99591 
99588 
99586 
99583 


I  09179 

i ! 09208 
i 109237 
'09266 


0915099580 


99578 
99575 
99572 
99570 


i  09295199567 
i  09324  89564 
: 09353 

j  09382|99559 
10941 1199556 
!  09440J99553 
1 09469199551 
09498J99548 
09527:99545 


59 
5H 
57 
56 
55 
54 
53 
52 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
33 
37 
36 
o5 
34 
33 
32 


Tana. 


I:09556|99542l31 
|  i  09585199540 
09614)99537 
!  109642199534 
'0967199531 
;  |09700i99528 
'09729  99626 
!' 03758  99523 
j  09787  99520 
!|  09816  99517 
1 1 09846(99614 
ji  09874 
i  i  09903:99508 
;  109932  9950'; 
l|  09961  (99503 
03990  99500 
10019J99497 
|  10048  99494 

ii  10077 

!  110103  99488 
|  10135  99485 
I  10164  99482 
I  10192)99479 

'110221  99470 
1025099473 
10279  99470 
I  10308  99467 
110337  99464 
10366  99401 
1 10398  99458 
110424  99455 
10453  99452 
N.  cos.  NVir>f. 


S4  Degree*. 


TABLE  II. 


Log.  Sines  and  Tangents.  (6°)  Natural  Sines. 


Cotang.  j  N.  sine.  N.  cos! 


Tang.  iD.  10'' 


021620 
022834 
024044 
025251 
026455 
027655 
028852 
030046 
031237 
032425 
033609 
,034791 
035969 
037144 
038316 
039485 
040651 
041813 
.042973 
044130 
045284 
, 046434 
047582 
048727 
049869 
051008 
052144 
053277 
054407 
055535 
056659 
,057781 
058900 
060016 
061130 
062240 
083348 
064453 
065556 
066655 
067752 
,068846 
069038 
071027 
072113 
073197 
074278 
075356 
076432 
077505 
078576 
.079644 
080710 
081773 
082833 
083891 
084947 
086000 
087050 
088098 
089144 


Cotang. 


202 
202 
201 
201 
200 
199 
199 
198 
198 
197 
197 
196 
196 
195 
195 
194 
194 
193 
193 
192 
192 
191 
191 
190 
190 
189 
189 
188 
188 
187 
187 
186 
186 
185 
185 
185 
184 
184 
183 
183 
182 
182 
181 
181 
181 
180 
180 
179 
179 
178 
178 
178 
177 
177 
176 
176 
175 
175 
175 
174 


10.9783801, 
977166  j 
975956 
97474911 
973545 | j 
972345  1 1 
971148  I; 
969954;! 
968763  !| 
967575  1 1 
966391 1 ! 

10.965209  ' 
964031 
962856 
961684'| 
960515 
959349 
958187 
957027 
955870 
954716 

10-953566 
952418 
951273 
950131 
94.8992 
947856 
946723 
945593 
944465 
943341 

10.942219 
941100 
939984 
938870 
937760 
936652 
935547 
934444 
933345 
932248 

10-931154 
930062 
928973 
927887 
926803 
925722 
924644 
923568 
922495 
921424 

10-920356 
919290 
918227 
917167 
916109 
915053 
914000 
912950 
911902  ! 
910856 

I   Tang! 


0453 
0482 
0511 
0540 
0569 
059; 
0626 
0655 
0684 
0713 
0742 
0771 
0800 
0829 
0858 
0887 
0916 
0945 
0973 
1002 
1031 
1060 
1089 
1118 
1147 
1176 
1205 
1234 
1263 
1291 
1320 
1349 
1378 
1407 
1436 
1465 
1494 
1523 
1552 
1580 


1638 

1667 

1696 

1725 

1754 

1783 

1812 

1840 

1869 

1898 

192' 

1956 

1985 

2014 

2043 

2071 

2100 

2129 

2158 

218 


99452 
>9449 
99446 
99443 
99440 
99437 
99434 
99431 
9y428 
99424 
99421 
99418 
99415 
99412 
99409 
99406 
99402 
99399 
99396 
99393 
99390 
99386 
99383 
99380 
99377 
H9374 
99370 
99367 
99364 
99360 
99357 
99354 
99351 
99347 
99344 
99341 
99337 
99334 
99331 
99327 


160999324 


99320 
99317 
99314 
99310 
99307 
99303 
99300 
99297 
99293 
99290 
99286 
99283 
99279 
99276 
99272 
99269 
992o5 
99262 
99258 
99266 


N.  ens.  N.sine 


83  Degrees. 


28 


Log.  Sines  and  Tangents.     (7°)     Natural  Sines. 


TABLE  II. 


Sine. 

9.035894 
086922 
087947 
038970 
089990 
091008 
092024 
093037 
094047 
095056 
096082 

9.097065 
098066 
099065 
100082 
101056 
102048 
103037 
104025 
105010 
105992 

9.106973 
107951 
108927 
109901 
110873 
111842 
112809 
113774 
114737 
115698 

9.116656 
117613 
118567 
119519 
120469 
121417 
122362 
123306 
124248 
125187 

9.126125 
127060 
127993 
128925 
129854 
130781 
131706 
132630 
133551 
134470 

9.135387 
136303 
137216 
138128 
139037 
139944 
140850 
141754 
142655 
143555 
Cosine 


D.  10' 


171 
171 
170 
170 
170 
169 
169 
168 
168 
168 
167 
167 
166 
166 
166 
165 
165 
164 
164 
164 
163 
163 
163 
162 
162 
162 
161 
161 
160 
160 
160 
109 
159 
159 
158 
158 
158 
157 
157 
157 
156 
156 
156 
155 
155 
154 
154 
154 
153 
153 
153 
152 
152 
152 
152 
151 
151 
151 
150 
150 


Cosine. 


D.  lu' 


.996751 
996735 
996720 
996704 
996688 
996673 
996657 
996641 
996625 
996610 
996594 
.996578 
996562 
996546 
996530 
996514 


996482 
996465 
996449 
996433 
.996417 
996400 
996384 
996368 
906351 
996335 
996318 
996302 
996285 
996269 
.996252 
996235 
996219 
996202 
996185 
996168 
996151 
996134 
996117 
996100 
.996083 
996066 
996049 
996032 
996015 
995998 
995980 
995963 
995946 
995928 
.995911 
995894 
995878 
995859 
995841 
995823 
995808 
995788 
995771 
995753 


2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.6 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.7 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.8 
2.9 
2.9 
2.9 
2.9 
2.9 


2.9 
2.9 
2.9 
2.9 
2.9 
2.9 
2.9 
2.9 


Sine. 


Taiiy;. 


.089144 
090187 
091228 
092266 
093302 
094336 
095367 
096395 
097422 
098446 
099468 

, 100487 
101504 
102519 
103532 
104542 
105550 
106556 
107559 
108560 
109559 

.110556 
111551 
112543 
113533 
114521 
115507 
116491 
117472 
118462 
119429 

. 120404 
121377 
122348 
123317 
124284 
125249 
126211 
127172 
128130 
129087 

.130041 
130994 
131944 
132893 
133839 
134784 
135726 
136667 
137605 
138542 

. 139476 
140409 
141340 
142269 
143196 
144121 
145044 
145966 
146885 
147803 


Cotang. 


D.  ju' 

174 
173 
173 
173 
172 
172 
171 
171 
171 
170 
170 
169 
169 
169 
168 
168 
168 
167 
167 
166 
166 
166 
165 
166 
165 
164 
164 
164 
163 
163 
162 
162 
162 
161 
161 
161 
160 
160 
160 
159 
159 
159 
158 
158 
158 
157 
157 
157 
166 
156 
156 
155 
155 
155 
154 
154 
154 
153 
153 
153 


uiaag.     kN.  sine.  N.  cos 


10.910356 
909813 
908772 
907734 
906698 
905664 
904633 
903605 
902578 
901554 
900532 

10.899513 
898496 
897481 
896468 
895458 
894450 
893444 
892441 
891440 
890441 

10.889444 
888449 
887457 
886467 
885479 
884493 
883509 
882528 
881548 
880571 

10.879596 
878623 
877652 
876683 
875716 
874761 
873789 
872828 
871870 
870913 

10.869959 
869006 
868056 
867107 
866161 
865216 
864274 
863333 
862395 
861458 

10.860524 
859591 
858660 
857731 
856804 
855879 
854956 
854034 
853115 
852197 


12187 
12216 
12245 
12274 
12302 
12331 
12360 
12389 
12418 
12447 
12476 
12504 
12533 
12562 
12591 
12620 
12649 
12678 
12708 
12735 
12764 
12793 
12822 
12851 
12880 
12908 
12937 
12966 
12995 
13024 
13053 
13081 
13110 
13139 
13168 
13197 
13226 
13254 
13283 
13312 
13341 
13370 


99255 
99251 
99248 
99244 
99240 
99237 
99233 
99230 
9226 
99222 
99219 
99215 
99211 
99208 
99204 
99200 
99197 
99193 
99189 
99180 
99182 
99178 
99175 
99171 
99167 
99163 
99160 
99156 
99152 
99148 
99144 
99141 
99137 
99133 
99129 
99125, 
99122" 
99118 
99114 
99110 
99106 
)9102 


13899  f99098 


13427 
13456 
13485 
13514 
13543 
13572 
13600 
13629 
13658 
13687 
13716 
13744 
13773 
13802 
13831 
13860 
13889 
1S917 


TllllL 


99094 
99091 

99087 
99083 
99079 
99076 
99071 
99067 
99063 
99059 
99055 
)9051 
J9047 
59043 
99039 
)9035 
J9031 
J9027 


60 

59 
58 
57 
56 
55 
64 
53 
62 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
35 
34 
33 
32 
31 
3D 
29 
28 
27 
26 
25 
24 
•,'3 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 
11 
10 
9 
S 
7 
6 
5 
4 
3 
2 
1 
0 


82  Degrees. 


30 


Log.  Sines  and  Tangents.     (9?)     Natural  Sines. 


TABLE  II. 


0 

1 

2 
3 
4 
5 
6 
7 
8 
9 
10 

11 

12 

13 
14 
15 
16 
V 
18 
19 
20 
•2! 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
53 
34 

35 


38 

39 
49 
41 
42 
43 
44 
45 
49 
47 
4o 
48 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 


Sine. 

.194332 
195129 
195925 
196/19 
197511 
198302 
199091 
199879 
200366 
201451 
202234 

.203017 
203797 
204577 
205354 
208131 
203906 
207679 
203452 
209222 
209992 

.210760 
211526 
212291 
213055 
213818 
214579 
215338 
216097 
216854 
217009 

.218363 
219116 
219868 
220318 
221307 
222115 
222861 
223600 
224349 
225092 

,225833 
226573 
227311 
228048 
228784 
229518 
230252 
230984 
231714 
232444 

.233172 
233899 
234625 
235349 
236073 
236795 
237515 
238235 
238953 
239070 


Cosine. 


I).  10'      Cosine.     D.  10 


133 
133 
132 
132 
132 
132 
131 
131 
131 
131 
130 
130 
130 
130 
129 
129 
129 
129 
128 
128 
128 
128 
127 
127 
127 
127 
127 
126 
126 
126 
126 
125 
125 
125 
125 
125 
124 
124 
124 
124 
123 
123 
123 
123 
123 
122 
122 
122 
122 
122 
121 
121 
121 
121 
120 
120 
120 
120 
120 
119 


'.994620 
994600 
994580 
994560 
994540 
994519 
994499 
994479 
994459 
994438 
991418 

'.994397 
99437/ 
994357 
994336 
994316 
994295 
994274 
994254 
994233 
994212 

.994191 
994171 
994150 
994129 
99410c 
994087 
994066 
994045 
994024 
994003 

.993981 
993960 
993939 
993918 
993896 
9938/5 
99385 i 
993832 
993811 
993789 

.993768 
993746 
993725 
993703 
993681 
993660 
993638 
993616 
993594 
993572 

.993550 
994528 
993506 
993484 
993462 
993440 
993418 

-993396 
993374 
993351 
Sine. 


3.3 
3.3 
3.3 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.4 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.5 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.6 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.7 


Tung. 

.199713 
20J529 
201345 
202159 
202971 
203782 
204592 
205400 
206207 
207013 
20/817 

.208619 
209420 
210220 
211018 
211815 
212611 
213405 
214198 
214989 
215780 

.216568 
217356 
218142 
218926 
219710 
220492 
221272 
222052 
222830 
223606 

^.224382 
225156 
225929 
226700 
227471 
228239 
229007 
229773 
230539 
231302 

.232065 
232826 
233586 
234345 
235103 
235859 
236614 
237368 
238120 
238872 

.239622 
240371 
241118 
241865 
242610 
243354 
244097 
244839 
245579 
246319 

Cotang. 


1).  10' 

136 
136 
136 
135 
135 
135 
135 
134 
134 
134 
134 
133 
133 
133 
133 
133 
132 
132 
132 
132 
131 
131 
131 
131 
130 
130 
130 
130 
130 
129 
129 
129 
129 
129 
128 
128 
128 
128 
127 
127 
127 
127 
127 
126 
126 
126 
126 
126 
125 
125 
125 
125 
125 
124 
124 
124 
124 
124 
123 
123 


Cotang.     i  IN.  sine.  N.  cos.l 


10.800287 
799471 
798655 
797841 
797029 
796218 
795403 
794600 
793793 
792987 
792183 

10.791381 
790580 
789780 
788982 
788185 
787389 
786595 
785802 
735011 
784220 

10.783432 
782644 
781858 
781074 
780290 
779503 
778728 
777948 
777170 
776394 

10.775618 
774844 
774071 
773300 
772529 
771761 
770993 
770227 
769461 
768698 

10.767935 
767174 
766414 
765655 
764897 
764141 
763386 
762632 
761880 
761128 

10.760378 
759629 
758882 
758135 
757390 
756646 
755903 
755161 
754421 
753681 


15643 
15672 
15701 
15730 
1575b 
15787 
15816 
15845 
15873 
15902 
15931 
15959 
159P8 
16017 
16046 
16074 
16103 
16132 
16160 
j 16189 
16218 
16246 


769 
93764 
98760 
98755 
93/51 
98746 
93741 
98737 
98732 
98728 
98723 
98718 
98714 
98709 
98704 
98700 
^8695 
98690 
98686 
98681 
98676 
98671 


16275  93667 

16304193662 

16333J98U57 

163;il98C,52 

1639098648 

16419  98643 

16447 198638 

||  16476  98633 

j!  16505  98629 

i!  16533198624 

!ll6562:98(il9 

ij  16591 198614 

1662098609 

i!  16648:98604 


1667  7 198600 


Tani 


j!  16708  98595 
116734^8690 

1 16763^8585 

j|  16792  98580 

ij  16820,98575 

I !  16849198570 

16878J98566 

1690698661 

1693698686 

16964  9:-;  551 

I  16992  98646 

li  17021  98541 

17050  98536 

ill7078|98531 

I!  17107  98626 

17136  98521 

1716498516 

17193  98511 

17222  98506 

17250  98501 

17279  98496 

17308  98491 

17336198486 

17365J98481 

V  cos.l N. sine, 


80  Dsgraas. 


TABLE  II. 


Log.  Sines  and  Tangents.     (10°)    Natural  Sines. 


31 


0  9 

i 


Sine. 

239670 

240385 
241101 
241814 
342526 

243037 
243947 
244656 
245363 
246069 
246775 

1.247478 
248181 
248883 
249583 
250282 
250980 
251677 
252373 
253037 
253761 

►  .254453 
255144 
255834 
256523 
257211 
257898 
258583 
259268 
259951 
200633 

'.261314 
261994 
262673 
263351 
264Q27 
264703 
265377 
266051 
266723 
267395 

L268065 
268  734 
269402 
270039 
270/35 
271400 
272064 
272726 
273388 
274049 

.274708 
275367 
276024 
276681 
277337 
277991 
278644 
279297 
279948 
280599 

Cosine. 


D.  lo"|    Cosine, 

1.993351 
993329 
993307 
993285 
993262 
993240 
993217 
993195 
993172 
993149 
993127 

'.993104 
993081 
993059 
993036 
993013 
992990 
992967 
992944 
992921 
992898 
•992875 
992852 
993829 
992806 
992783 
992759 
992736 
992713 
992690 
992666 

.992643 
992619 
992596 
992572 
992549 
992525 
992501 
992478 
992454 
992430 

.992406 
992382 
992359 
992335 
992311 
992287 
992263 
992239 
992214 
992190 

.992166 
992142 
992117 
992093 
992069 
992044 
992020 
991996 
991971 
991947 


119 
119 
119 
119 
118 
118 
118 
118 
118 
117 
117 
117 
117 
117 
116 
116 
116 
116 
116 
116 
115 
115 
115 
115 
115 
114 
114 
114 
114 
114 
113 
113 
113 
113 
113 
113 
112 
112 
112 
112 
112 
111 
111 
111 
111 
111 
111 
110 
110 
110 
110 
110 
110 
109 
109 
109 
109 
109 
109 
108 


Sine. 


D.  W 

3.7 
3.7 
3.7 
3.7 
3.7 
3.7 
3.8 
3.8 
3.8 
3.8 
3.8 
3.8 
3.8 
3.8 


3 

3 

3 

3 

3 

3 

3, 

3.8 

3.8 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

3.9 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 

4.0 


4.1 
4.1 


Tan' 


,246319 
247057 
247794 
248530 
249264 
249998 
250730 
251461 
252191 
252920 
253648 

.254374 
255100 
255824 
256547 
257269 
257990 
258710 
259429 
260146 
260863 

.261578 
262292 
263005 
263717 
264428 
265138 
265847 
266555 
267261 
267967 

.268671 
269375 
270077 
270779 
271479 
272178 
272876 
273573 
274269 
274964 

.275658 
276351 
277043 
277734 
278424 
279113 
279801 
280488 
281174 
281858 

.282542 
283225 
283907 
284588 
285268 
285947 
286624 
287301 
287977 
288652 


D.  10"|  Cotang. 


123 
123 
123 
122 
122 
122 
122 
122 
121 
121 
121 
121 
121 
120 
120 
120 
120 
120 
120 
119 
119 
119 
119 
119 
118 
118 
118 
118 
118 
118 
117 
117 
117 
117 
117 
116 
116 
116 
116 
116 
116 
115 
115 
115 
115 
115 
115 
114 
114 
114 
114 
114 
114 
113 
113 
113 
113 
113 
113 
112 


Cotang. 
Degrees. 


IN.sine.lN.  cos, 


10.753681 
75v943 
75^205 
751470 
750736 
750002 
749270 
748539 
747809 
747080 
746352 

10.745626 
744900 
744176 
743453 
742731 
742010 
741290 
740571 
739854 
739137 

10.738422 
737708 
736995 
736283 
735572 
734862 
734153 
733445 
732739 
732033 

10.731329 
730625 
729923 
729221 
728521  jj 
727822 
727124 
726427 
725731 
725036 
724342 
723649 
722957 
722266 
721576 
720887 
720199 
719512 
718826 
718142 

10.717458 
716775 
716093 
715412 
714732 
714053 
713376 
712699 
712023 
711348 


.7365198481 


10 


Tang. 


98476 
98471 
98466 
98461 
98455 
98450 
98445 
98440 
98435 
98430 
98425 
98420 
98414 
98409 
98404 
98399 
983y4 
98389 
98383 
98378 
98373 
98368 
98362 
98357 
98S52 
98347 
98341 
98336 
98331 
98325 
18252.98320 
18281 '983  J  5 
18309!983l0 
IS338:9ho04 
18367198299 
1S395I98294 
1842498288 
18452  98283 


17393 
17422 
17451 
17479 
17508 
17537 
17565 
17594 
17623 
17651 
17680 
17708 
17737 
17766 
17794 
17823 
17852 
17880 
17909 
17937 
17966 
17995 
18023 
18052 
18081 
18109 
18138 
18166 
18195 
18224 


18481 
18509 
18538 
18567 
18595 
18624 
18652 
18681 


98277 
98272 
98267 
98261 
98256 
98250 
98245 
98240 


18710  98234 
18738;98229 
18  767 198223 
18795J98218 
1882498212 
1885298207 
1888198201 
1891098196 
18938  98190 
18967;98185 
18995i'98179 
19024I98174 
19052198168 
19081  [98163 

N.  cos.  N.s-ine. 


>  and  Tangents.  (11°)  Natural  Bines. 


TABLE  II. 


Sin. 


ID.  10 


1 

2 
3 

4 

5 

6 

7 

4 

9 

10 

11 

12 

13 

14 

15 

16 

r 

18 

19 
20 

21 
22 
23 

24 
20 
2G 

2; 

24 
•J') 
30 
31 
32 
33 
34 
30 
36 
3^ 
38 
39 
40 
41 
42 
43 
41 
45 
46 
4? 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
53 
59 
60 


.280599 
281248 
281897 
282544 
283190 
283836 
284480 
285124 
285766 
286408 
287048 
.287687 
288326 
288964 
289600 
290236 
290870 
291504 
292137 
292768 
293399 
.294029 
294658 
295286 
295913 
296539 
29/164 
297788 
298412 
299034 
299655 
.300276 
300895 
301514 
302132 
302748 
303364 
303979 
304593 
305207 
305819 
.306430 
307041 
307650 
308259 
308867 
309474 
310080 
310685 
311289 
311893 
.312495 
313097 
313698 
314297 
314897 
315495 
316092 
316689 
317284 
317879 
('(.•sine 


103 
103 
108 
108 
108 
107 
107 
107 
107 
107 
107 
106 
106 
106 
106 
106 
106 
105 
105 
105 
105 
105 
105 
104 
104 
104 
104 
104 
104 
104 
103 
103 
103 
103 
103 
103 
102 
102 
102 
102 
102 
102 
102 
101 
101 
101 
101 
101 
101 
100 
100 
100 
100 
100 
100 
100 
100 
99 
99 
99 


ID.  i, 


.991947 
991922 
991897 
991873 
991848 
991823 
991799 
991774 
991749 
991724 
991699 

.991674 
991649 
991824 
991599 
991574 
991549 
991524 
991498 
991473 
991448 

.991422 
991397 
991372 
991346 
991321 
991295 
991270 
991244 
991218 
991193 

.991167 
991141 
991115 
991090 
991064 
991038 
991012 
990986 
990960 
990934 
990908 
990842 
990855 
990829 
990803 
990777 
990750 
990724 
690697 
990671 
.990644 
990618 
990591 
990565 
990538 
990511 
990485 
990458 
990431 
990404 


Sine. 


4.1 
4.1 
4.1 
4.1 
4.1 
4.1 
4.1 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.2 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.3 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.4 
4.5 
4.5 
4.5 
4.5 


,288652 
24932  0 
289999 
290671 
291342 
292013 
292682 
293350 
294017 
294684 
295349 

.298013 
298677 
297339 
298001 
298662 
299322 
299980 
300638 
301295 
301951 

.302607 
303261 
303914' 
304567 
305218 
305869 
306519 
307168 
307815 
308463 

.309109 
309754 
310398 
311042 
311685 
312327 
312967 
313608 
314247 
314885 

.315523 
316159 
316795 
317430 
318064 
318697 
319329 
319961 
320592 
321222 

.321851 
322479 
323106 
323733 
324358 
324983 
325607 
326231 
326853 
327475 


D.    IK) 

112 
112 
112 
112 
112 
111 
111 
111 
111 
111 
111 
111 
110 
110 
110 
110 
110 
110 
109 
109 
109 
109 
109 
109 
109 
108 
108 
108 
108 
108 
108 
107 
107 
107 
107 
107 
107 
107 
106 
106 
106 
106 
106 
106 
108 
105 
105 
105 
105 
105 
105 
105 
104 
104 
104 
104 
104 
104 
104 
104 


Cotan<>;. 


10 


10.711344 
71<J674 
710001 
709329 
708658 
707947 
707318 
706650 
705983 
705316 
704651 
■  702987 
703323 
702661 
701999 , 
701338 ! 
700678 
700020 
699362 
698705 
698049 

10-697393 
696739 
696086 
695433 
694782 
694131 
693481 
692832 
692185 
691537 

10-690891 
690246 
689602 
688958 
688315 
687673 
687033 
686392 
'  685753 
685115 

10-684477 
683841 
683205 
682570 
681936 
681303 
680671 
680039 
679408 
678778 

10-678149  I 
677521 
676894 I 
676267  ; 
675642 
676017 
674393 
673769 
673147 
672525 I 


19138 
19167 
19195 
19224 
19252 
19281 
19309 
19338 
19366 
19395 
19423 
19452 
19481 
19509 
19538 
19566 
19595 
19623 
19652 
19680 
19709 
19737 
19766 
19794 
19823 
19851 
19880 
19908 
19937 
19965 
19994 
20022 
20051 
20079 
20108 
20136 
20165 
20193 
20222 
20250 
20279 
20305 
20336 
20364 
20393 
120421 
! 20450 
1 20478 
;  20507 
120535 
•  20563 
20592 
1 20620 
J  20649 
20677 
20?  06 
!  20734 
j 20763 
'20791 


98152 
94146 
94140 
98135 
98129 
98124 
98118 
98112 
98107 
98101 
98096 
98090 
98084 
98079 
98073 
98067 
98061 
98056 


98050  40 


Tan: 


i|  X.  cos.  N-aine. 


98044 
98039 
98033 
98027 
98021 
98016 
98010 
98004 
97998 
97992 
97987 
97981 
97975 
97969 
97963 
97958 
97952 
97946 
97940 
97934 
97928 
97922 
97916 
97910 
97905 
97899 
97893 
97887 
97881 
97875 
97869 
9)863 
97857 
97851 
7845 
97839 
97833 
97827 
97821 
97815 


39 
38 
37 
36 
35 
34 
33 
32 
31 
30 
29 
28 
27 
26 
25 
24 
28 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 
11 
10 
9 
4 
7 
6 
6 
4 

3 
2 
1 
0 


78  Decrees. 


TABLE  II.  Log.  Sines  and  Tangents.    (12°)    Natural  Sines. 


33 


11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
2-2 
23 
24 

26 
27 

28 
29 
30 

31 
82 
33 
34 
I  35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
46 
49 
50 
61 
52 
53 
64 
55 
56 
5  7 
68 
59 
60 


Sine. 

.317879 
318473 
319  )i. 
319658 
3-20249 
329840 
321430 
322019 
322607 
323194 
323780 
.324366 
324950 
325534 
326117 
326709 
327281 
327862 
3284.42 
32;)J21 
329599 
.330176 
330753 
331329 
331903 
332478 
333051 
333624 
334195 
334766 
335337 
.335903 
336475 
337043 
337610 
338176 
338742 
339306 
339871 
340434 
340996 
,341558 
342119 
342679 
343239 
343797 
344355 
344912 
345469 
346024 
346579 
,347134 
347687 
348240 
348792 
349343 
349893 
3 j 0443 
350992 
351540 
352088 


L).  10' 


99.0 
93.8 
98.7 
98.6 
98.4 
98.3 
98.2 
98.0 
97.9 
97.7 
97.6 
97.5 
97.3 
97.2 
97.0 
96.9 
96.8 
96.6 
96.5 
96.4 
96.2 
96.1 
96.0 
95.8 
95.7 
95.6 
95.4 
95.3 
95.2 
95.0 
94.9 
94.8 
94.6 
94.5 
94.4 
94.3 
94.1 
94.0 


Cosine. 


5 
4 

2 
1 

93.0 
92.9 
92.7 
92.6 
92.5 
92.4 
92.2 
92.1 
92.0 
91.9 
91.7 
91.6 
91.5 
91.4 
91.3 


Oesine. 

.990404 
990378 
999351 
999324 
990297 
993270 
990243 
999215 
990188 
999161 
999134 
.999107 
990079 
990052 
999925 
989997 
989970 
989942 
989915 
989887 
989860 
.989832 
939804 
989777 
989749 
989721 
989893 
989665 
989837 
989609 
989582 
.989553 
989525 
989497 
989469 
989441 
989413 
989384 
989356 
989328 
989300 
.989271 
989243 
989214 
989186 
989157 
989128 
989100 
989071 
989042 
989014 
.988985 
988956 
988927 


988869 
988840 
988811 
988782 
988753 
988724 


Sine. 


L>.  10 


4.5 
4.5 
4.5 
4.5 
4.5 
4.5 
4.5 
4.5 
4.5 
4.5 


4 
4 
4 

4 
4 

4 

4 

4 

4.6 

4.6 

4.6 

4.6 

4.6 

4.6 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.7 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.8 

4.9 

4.9 

4.9 


9.327474 
328095 
328715 
329334 
329953 
330570 
331187 
331803 
332418 
333033 
333646 
334259 
334871 
335482 
336093 
336702 
337311 
337919 
338527 
339133 
339739 
340344 
340948 
341552 
342155 
342757 
343358 
343958 
344558 
345157 
345755 

9.346353 
346949 
347545 
348141 
348735 
349329 
349922 
350514 
351106 
351697 
352287 
352876 
353465 
354053 
354640 
355227 
355813 
356398 
356982 
357566 

9.358149 
358731 
359313 
359893 
360474 
361053 
361632 
362210 
362787 
363364 


Cotang. 


0.  lu 

103 
103 
103 
103 
103 
103 
103 
102 
102 
102 
102 
102 
102 
102 
102 
101 
101 
101 
101 
101 
101 
101 
101 
100 
100 
100 
100 
100 
100 
100 
100 
99. 
99. 
99, 
99, 
99, 


Cocang. 

10.672526 
671905 
671285 
670666 
670047 
689430 
668813 
668197 
667582 
666967 
666354 

10.665741 
665129 
664518 
663907 
663298 
662689 
662081 
661473 
660867 
660261 

10.659656 
659052 
658448 
657845 
657243 
656642 
656042 
655442 
654843 
654245 

10.653647 
653051 
652455 
651859 
651265 
650671 
650078 
649486 
648894 
648303 

10.647713 
647124 
646535 
645947 
645360 
644773 
644187 
643602 
643018 
642434 

10.641851 
641269 
640687 
640107 
639526 
638947 
638368 
637790 
637213 
636636 


N.  sine.  iN .  cos. 


20791  97815 

20820  97809 

20848  97803 

20877  97797 

20905  97791 

20933  97784 

20962  97778 

20990  97772 

21019  97766 

21047  97760 

21076  97754 

21104  9774- 

21132  97742 

2116197735 

21189  97729 

21218  97723 

21246  97717 

21275  97711 

21303  97705 

21331  97698 

21360  97692 

21388  97686 

21417  97689 

21445  97673 

21474  97667 

21502  97661 

2153097655 

21559  97648 

ii  21587  97642 

21616  97636 

21644  97630 

21672  97623 

21701  ;976l7 

21729  97611 

j|  21758  97604 

21786  97598 

II 21814  97592 

i!21843:97585 

1 21871 197579 

!21899!97573 

1 121928 197566 

1121956  97560 

!  |2198597553 

1 122013197547 

,'22041:97541 

122070:97534 

122098  97528 

| !  22126  97521 

i|  22155  97515 

I' 22183 '97508 

122212  97502 

1 22240|97496 

i  22268  97489 

22297197483 

22325197476 

22353|97470 

2238297463 

22410197457 

22438  97450 


22467 
22495 


Tang. 


N.  cos.  N.sine. 


97444 
97437 


77  Degrees. 


Log.  Sines  and  Tangents.  (13°)  Natural  Sines. 


TABLE  II. 


Sine.  D.  1U"|  Cosine. 


0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 

11 

1-2 
13 

14 

15 

lb 
1? 

IS 
19 
20 
21 

■>■: 

23 

24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
3(3 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
61 
62 
53 
54 
55 
56 
67 
58 
59 
GO 


.352088 
352635 
353181 
353726 
354271 
354815 
355358 
355901 
356443 
356984 
357524 

.358064 
358603 
359141 
359578 
360215 
360752 
361287 
361822 
362356 
3W889 

.363422 
363954 
364485 
365016 
365546 
366075 
366604 
367131 
367659 
368185 

.368711 
369236 
369761 
370285 
370808 
371330 
371852 
372373 
372894 
373414 

. 373933 
374452 
3749/0 
375487 
376003 
376519 
377035 
377549 
378063 
378577 

.379089 
379601 
380113 
380624 
381134 
381643 
382152 
382661 
383168 
383675 
Cosine. 


91.1 

91.0 

90.9 

90.8 

90.7 

90.5 

90.4 

90.3 

90.2 

90.1 

89.9 

89.8 

89.7 

89.6 

89.5 

89.3 

89.2 

89.1 

89.0 

88.9 

88.8 

88.7 

88.5 

88.4 

88.3 

88.2 

88.1 

88.0 

87.9 

87.7 

87.6 

87.5 

87.4 

87.3 

87.2 

87.1 

87.0 

86.9 

86.7 

86.6 

86.5 

86.4 

86.3 

86.2- 

86.1 

86.0 

85.9 

85.8 

85.7 

85.6 

85.4 

85.3 

§5.2 

85.1 

85.0 

84.9 

84.8 

84.7 

84.6 

84.5 


.988724 
938695 
988666 
988636 
988607 
988578 
988548 
988519 
988489 
988460 
988430 
.988401 
988371 
988342 
988312 
988282 
988252 
988223 
988193 
988163 
988133 
.988103 
988073 
988043 
988013 
987983 
987953 
987922 
987892 
987862 
987832 
.987801 
987771 
987740 
987710 
987679 
987649 
987618 
987588 
987557 
987526 
.987496 
987465 
987434 
987403 
987372 
987341 
987310 
987279 
987248 
987217 
'.987186 
987155 
987124 
987092 
987061 
987030 
986998 
986967 
986936 
986904 


Sine. 


).  10' 

4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
6.0 
5.0 
6.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.1 
6.1 
5.1 
5.1 
5.1 


1 
1 
1 

2 
2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
6.2 
5.2 
5.2 
5.2 
5.2 
5.2 
6.2 
6.2 


I'm;. 


9.363364 
363940 
364515 
365090 
365664 
366237 
366810 
367382 
367953 
368524 
369094 

9.369663 
370232 
370799 
371367 
371933 
372499 
373084 
373629 
374193 
374756 
375319 
375881 
376442 
377003 
377563 
378122 
378681 
379239 
379797 
380354 

9.380910 
381466 
382020 
382575 
383129 
383682 
384234 
384786 
385337 
385888 
386438 
386987 
387536 
388084 
388631 
389178 
389724 
390270 
390815 
391360 

9.391903 
392447 
392989 
393531 
394073 
394614 
395154 
395694 
396233 
396771 


Cotam 


D.  10"|  (Jotanj*.  i  N.sine  -N.  cos, 


10 


10 


10 


10 


10 


10 


636636 
636060 
635485 
634910 
634336 
633763 
633190 
632618 
632047 
631476 
630906 
630337 
629768 
629201 
628633 
628067 
627501 
626936 
626371 
625807 
625244 
624681 
624119 
623558 
622997 
622437 
621878 
621319 
620761 
620203 
619646 
,619090 
618534 
617980 
617425 
616871 
616318 
615766 
615214 
614663 
614112 
,613562 
613013 
612464 
611916 
611369 
610822 
610276 
609730 
60J185 
603640 
.608097 
607553 
607011 
606469 
605927 
605386 
604846 
604306 
603767 
603229 
TangT 


i  22495 
: 22523 
| 22552 
! 22580 
| 22608 
1 22637 
! 22665 
| 22693 
! 22722 


22750  97378 


22778 
22807 
22835 


22977 
23005 
23033 
23062 


97437 
97430 
97424 
97417 
97411 
97404 
97398 
97391 
97384 


9737  M 
97365 
97358 


22863'97351 
22892197345 
22920  973L-8 
22948  97331 


97325 
97318 
97311 
97304 


23090  97298 


23118 
23146 
23175 
23203 
23231 
23260 
23288 
23316 
23345 
23373 
23401 
23429 
23458 
23486 
23514 
23542 
23571 
23599 
23627 
23656 
23684 
23712 


23740  97141 


23769 
23797 
23825 


23938 


97291 
97*84 
97278 
97271 
97264 
9/257 
97251 
97244 
97237 
97230 
97223 
97217 
97210 
97203 
97  96 
97189 
;>7182 
97176 
97169 
97162 
97155 
97148 


97134 
97127 
97120 


23853  97113 
23882  97103 
23910  97100 


•J7093 


23966  9  70^6 
23995  97079 
24023  970/2 
24051  97065 
24079  97053 
24108  97051 
2413b  97044 
24164  97037 
2419-2  97030 


n.  cos.  |n. 


\Q  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (14°)    Natural  Sines. 


35 


9 

10 

11 

12 

13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
SO 
31 
32 
33 
34 
35 
86 
3? 
3d 
30 
40 
41 
42 
40 
44 
45 
46 
47 
43 
40 
50 
51 
52' 
53 
54 
55 
50 
57 
58 
50 
60 


9. 


.333675 
384182 
384687 
385102 
385697 
380201 
386704 
38720/ 
387709 
388210 
388711 
.389211 
389711 
390210 
390708 
391206 
391703 
392199 
392695 
393191 
393685 
.394179 
394673 
395166 
395658 
396150 
396641 
397132 
397621 
398111 
398000 
.399088 
399575 
400062 
400549 
401035 
401520 
402005 
402489 
402972 
403455 
403938 
404420 
404901 
405382 
405862 
406341 
406820 
407299 
407777 
408254 
408731 
409207 
409682 
410157 
410632 
411100 
411579 
412052 
412524 
412996 
Cosine. 


L>.  lu' 


84.4 

84.3 

84.2 

81.1 

81.0 

83.9 

83.8 

83.7 

S3. 6 

83.5 

33.4 

83.3 

83.2 

83.1 

83.0 

82.8 

82.7 

82.6 

82.5 

82.4 

82.3 

82.2 

82.1 

82.0 

81.9 

81.8 

81.7 

81.7 

81,6 

81.5 

81.4 

81.3 

81.2 

81.1 

81.0 

80.9 

80.8 

80.7 

80.6 

80.5 

80.4 

80.3 

80.2 

80.1 

80.0 

79.9 

79.8 

79.7 

7y.6 

79.5 

79.4 

79.4 

79.3 

79.2 

79.1 

79.0 

78.9 

78.8 

78.7 

78.6 


9. 


.986904 
986873 
986841 
986809 
986778 
936746 
986714 
986683 
986651 
980619 
986587 
.986555 
986523 
986491 
986459 
986427 
986395 
986363 
986331 
986299 
986266 
.986234 
986202 
986169 
986137 
986104 
986072 
986039 
986007 
985974 
985942 
985909 
985876 
985843 
985811 
985778 
985745 
985712 
985679 
985646 
985613 
985580 
985547 
985514 
985480 
985447 
985414 
985380 
985347 
985314 
985280 
985247 
985213 
985180 
985146 
985113 
985079 
985045 
985011 
984978 
984944 


D.  lu; 


Sim 


5.2 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.3 
5.4 
5.4 
5.4 
5.4 
5.4 
6.4 
6.4 
5.4 
5.4 
5.4 
5.4 
5.4 
5.4 
5.4 
5.5 
5.5 
5.5 
5.5 
5.5 
5.5 
5.6 
5.6 
5.5 
5.5 
5.5 
5.5 
5.5 
5.6 
5.5 
5.6 
5.6 
5.6 
5.6 


Taas 


5.6 
5.6 
5.6 

5.6 


9.396771 
397309 

397846 
398383 
398919 
399455 
399990 
400524 
401058 
401591 
402124 

9.402656 
403187 
403718 
404249 
404778 
405308 
405836 
406364 
406892 
407419 

9.407945 
408471 
408997 
409521 
410045 
410569 
411092 
411615 
412137 
412658 
413179 
413699 
414219 
414738 
415257 
415775 
416293 
416810 
417326 
417842 

9.418358 
418873 
419387 
419901 
420415 
420927 
421440 
421952 
422463 
422974 

9.423484 
423993 
424503 
425011 
425519 
426027 
426534 
427041 
427547 
428052 

Co  tang. 


D.  10'1 


89.6 

89.6 

89.5 

89.4 

89.3 

89.2 

8.9.1 

89.0 

88.9 

88.8 

88.7 

88.6 

88.5 

88.4 

88.3 

88.2 

88".  1 

88.0 

87.9 

87.8 

87.7 

87.6 

87.5 

87.4 

87.4 

87.3 

87.2 

87.1 

87.0 

86.9 

86.8 

86.7 

86.6 

86.5 

86.4 

86.4 

86.3" 

86.2 

86.1 

86.0 

85.9 

85.8 

85.7 

85.6 

85.5 

85.5 

85.4 

85.3 

85.2 

85.1 

85.0 

84.9 

84.8 

84.8 

84.7 

84.6 

84.5 

84.4 

84.3 

84.3 


Cotang.   X.  sin-\  N.  cos. 


10.603229!  124192 
602691  1 1 24220 
602154  ; 24249 
601617  1 1 24277 
601081!!  24305 


030 
97023 
97015 
97003 
97001 


600545 
6000101 
59J476 | 
598942 i 
598409 ! 
597876 ! 

10.597344! 
596813  ! 
596282 ! 
595751  I 
595222 : 
594692 ! 
504164  i 
5936361 

•  593108 | 
592581 

10.592055 


24333:90994 


24302 
24390 
24418 
24446 


96987 
96980 
96973 
96906 


24474^6959 


»£ 


591529  |  24813 


24503 
24531 
24559 
24587 
24615 
24644 
24672 
24700 
24728 
24756 
24784 


691003 
590479 ! 
589955  j 
589431 ! 
588908 
588385 
687863 
587342 

10.586821 
586301 
585781 
585262 
584743 
584225 
583707 
583190 
582674 
582158 

10.581642 
581127 
680613 
680099 
679585 
579073 ! 
678560 ! 
578048 ! 
577537  I 
577026  I 

10.576516 
576007 
575497 


24841 
24869 


24897  96851 


24925 
i  24954 
■'ll  24982 
25010 
25038 
25006 
25094 
25122 
25151 
25179 
25207 
25235 
25263 
25291 
25320 
25348 
25376 
25404 
25432 
25460 
25488 
25516 
25545 
25573 
25601 
25629 
25657 
j 25685 
574989!  25713 
574481!  25741 
573973!  25766 
673466!  25798 
572959 !  25826 
572453  1 1 25854 
571948;  1 25882 
"Tang.   |  IN 


96952 
96945 
96937 
96930 
96923 
96916 
96909 
96902 
96894 
96887 
96880 
96873 
96866 
96858 


96844 
96837 
96829 
96822 
96815 
96807 
96800 
96793 
96786 
96778 
96771 
96764 
96756 
96749 
96742 
96734 
96727 
96719 
96712 
96705 
96697 
96690 
96682 
96675 
96667 
96660 
96653 
96645 
96638 
96630 
96623 
96615 
96608 
96600 
96593 
N.nlne. 


75  Degrees. 


20 


26 


Log.  Sines  and  Tangents.     (15°)     Natural  Sines. 


TABLE  II. 


bine. 

9.412993 
413467 
413938 
414408 
414878 
415347 
415815 
416283 
416751- 
417217 
417684 

9.418150 
418615 
419079 
419544 
420007 
420470 
420933 
421395 
421857. 
422318 

9.422778 
423238 
4236*7 
424158 
424615 
425073 
425530 
425987 
426443 
426899 

9.427354 
427809 
428263 
428717 
429170 
429823 
430075 
430527 
430978 
431429 

9.431879 
432329 
432778 
433226 
433675 
434122 
434569 
435016 
435462 
435908 
436353 
436798 
437242 
437686 
438129 
438572 
439014 
439456 
439897 
440338 


D.  1U"| 

78.5 
78.4 
78.3 
78.3 

re.s 

78-1 
78.0 


a  .a 

77.8 
77.7 
77.6 
77.5 
77.4 
77.3 
77.3 
77.2 
77.1 
77.0 
76.9 
76.8 
76.7 
76.7 
76.6 
76.5 
76.4 
76.3 
76.2 
76.1 
76.0 
76.0 
75.9 
75.8 
75.7 
75.6 


76 

75 

75 

75 

75 

75 

75 

74.9 

74.9 

74.8 


74 

74 

74 

74 

74 

74 

74 

74.1 

74.0 

74.0 

73.9 

73.8 

73.7 

73.6 

73.6 

73.5 


Cosine,    j 


.984944 
984910 
984876 
984842 
984808 
984774 
984740 
984706 
984672 
984637 
984603 

.984569 
984535 
984500 
981466 
984432 
984397 
984363 
984328 
984294 
984259 

.984224 
984190 
984155 
984120 
984085 
984050 
984015 
983981 
983946 
983911 

.983875 
983840 
983805 
983770 
983735 
983700 
983664 
983629 
983594 
983558 

.983523 
983487 
983452 
983416 
983381 
983345 
983309 
983273 
983238 
983202 

.983166 
983130 
983094 
983058 
983022 
982986 
982950 
982914 
982878 
982842 


D.  10' 

5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.7 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.8 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
5.9 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 
6.0 


9.428052 
428557 
429082 
429566 
430070 
430573 
431075 
431577 
432079 
432580 
433080 
433580 
434080 
434579 
435078 
435576 
436073 
436570 
437067 
437563 
438059 

9.438554 
439048; 
439543 
440036 
440529 
441022 
441514 
442008 
442497 
442988 

9.443479 
443968 
444458 
444947 
445435 
445923 
446411 
446898 
447384 
447870 
.448356 
448841 
449326 
449810 
450294 
450777 
451260 
451743 
452225 
452706 
.453187 
453668 
454148 
454628 
455107 
455586 
456064 
456542 
457019 
457496 
Co  tang. 


D.  10" 


N.  sin 


10.571948 
571443 
570938 
570434 
569930 
569427 
538925 
568423 
567921 
567420 
566920 

10.566420 
565920 
565421 
584922 
564424 
563927 
563430 
562933 
562437 
561941 

10.561446 
580952 
560457 
559964 
559471 
558978 
558486 
557994 
557503 
557012 

10.556521 
556032 
555542 
555053 
554565 
554077 
553589 
553102 
552616 
552130 

10.551644 
551159 
550674 
550190 
549706 
549223 
548740 
548257 
547775 
547294 

10.546813 
546332 
545852 
545372 
544893 
544414 
543936 
543458 
542981 
542504 


25882 
25910 
2593- 
2596o 
25994 
26022 
26050 
26079 
26107 
26135 
26163 
26191 
26219 
|26247 
i  26275 
| 26303 
| 26331 
j  26359 
! 26387 
26415 
26443 
26471 
26500 
26528 
26556 
26584 
26612 
26640 
26668 
26696 
26724 
I 26752 
26780 
; 26808 
26836 
! 26864 
| 26892 
26920 
26948 
26976 
27004 
27032 
27060 
27088 
27116 
27144 
27172 
27200 
27228 
27256 
27 

27312 
27340 
27368 
27396 
27424 
27452 
274S0 
27508 
27536 
27564 


96593 
96585 
96578 
96570 
96562 
96555 
96547 
96540 
96532 
96524 
96517 
96509 
96502 
96494 
96486 
98479 
96471 
96463 
96456 
96448 
96440 
96433 
96425 
96417 
96410 
9ti402 
96394 
96386 
96379 
96371 
96363 
96355 
96347 
96340 
96332 
96324 
96316 
96308 
96301 
96293 
96285 
96277 
96269 
96261 
96253 
96246 
96238 
96230 
96222 
96214 
284  96206 
96198 
96190 
96182 
96174 
96166 
96158 
96150 
96142 
96134 
96126 


Tang.   |  N.  coP.jN.gme, 


74  Degrees. 


TABLE  II. 


Lou'.  Sinefl  and  Tangents.    (16°)    Natural  Sines. 


37 


0 

3.410338 

1 

440778 

.» 

441218 

fl 

441058 

4 

44-20i)0 

5 

442535 

0 

442973 

7- 

443410 

8 

443847 

9 

444284 

10 

444720 

11 

9.445155 

1-2 

445590 

13 

446025 

14 

446459 

16 

446893 

16 

447326 

17 

447759 

18 

448191 

19 

448623 

•20 

449054 

21 

9.449485 

22 

449915 

2:; 

450345 

24 

450775 

25 

451204 

26 

451632 

•27 

452060 

28 

452488 

29 

452915 

80 

453342 

81 

9.453768 

32 

454194 

83 

454619 

34 

455044 

85 

455469 

80 

455893 

87 

456316 

88 

456739 

89 

457162 

40 

457584 

41 

9.458006 

42 

458427 

48 

458848 

44 

459268 

45 

459688 

40 

460108 

47 

460527 

48 

460946 

49 

461364 

50 

461782 

51 

9.462199 

52 

462616 

58 

463032 

54 

'  463448 

55 

463864 

50 

464279 

57 

464694 

58 

465108 

5!) 

466522 

00 

465935 

Cosine. 


I),  lo' 


73.4 
73.3 
73.2 
73-1 
73.1 
73.0 
72.9 
72.8 
72.7 
72.7 
72.6 
72.5 
72.4 
72.3 
72.3 
72.2 
72.1 
72.0 
72.0 
71.9 
71.8 
71.7 
71.6 
71.6 
71.5 
71.4 
71.3 
71.3 
71.2 
71.1 
71.0 
71.0 
70.9 
70.8 
70.7 
70.7 
70.6 
70.5 
70.4 
70.4 
70.3 
70.2 
70.1 
70.1 
70.0 
69.9 
69.8 
69.8 
69.7 
69.6 
69.5 
69.5 
69.4 
69.3 
69.3 
69.2 
69.1 
69.0 
69.0 
68.9 


.982842 
982805 
982709 
982733 
982690 
982660 
982624 
982587 
982551 
982514 
982477 

.982441 
982404 
982367 
982331 
982294 
982257 
982220 
982183 
982146 
982109 

.982072 
982035 
981998 
981961 
981924 
981886 
981849 
981812 
981774 
981737 

.981699 
981662 
981625 
981587 
981549 
981512 
981474 
981436 
981399 
981361 

.981323 
981285 
981247 
981209 
981171 
981133 
981095 
981057 
981019 
989^81 

.980942 
980904 
980866 
980827 
980789 
980750 
980712 
980673 
980635 
980596 
Sine. 


D.  10" 


6.0 
0.0 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.1 
6.2 
6.2 
6.2 
6.2 
6.2 
6.2 
6.2 


9.457496 


6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6.3 

6, 

0 

0 

0 

0 

6 

6 

6 

0 

0 

0 

6 

(i 

0 


3 
8 
3 
3 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
6.4 
6.4 
6.4 
6.4 


45 

45 

4589 

4594 

4598 


458449 


459400 


73 
49 
i25 
00 
75 
460349 
460823 
461297 
461770 
462242 
9  462714 
463186 
463658 
464129 
464699 
465069 
465539 
466008 
466476 
466945 
467413 
467880 
468347 
468814 
469280 
469746 
470211 
470676 
471141 
471605 
9  472068 
'472532 
472995 
473457 
473919 
474381 
474842 
475803 
475763 
476223 
476683 
477142 
477601 
478059 
478517 
478975 
479432 
479889 
480345 
480801 
481257 
481712 
482167 
482621 
483075 
483529 
483982 
484435 
484887 
485339 


9. 


Cotang. 


D.  10" 

79.4 
79.3 
79.3 
79  2 
79.1 
79.0 
79.0 
78.9 
78.8 
78.9 
78.7 
78.6 
78-5 
78.5 
78.4 
78.3 
78.3 
78.2 
78.1 
78.0 
78.0 
77-9 
77-8 
77.8 
77.7 
77-6 
77-5 
77.5 
77-4 
77-3 
77-3 
77.2 
77.1 
77.1 
7770 
76.9 
76.9 
76.8 
76.7 
76.7 
76.6 
76.5 
76.5 
76.4 
76.3 
76.3 
76.2 
76.1 
76.1 
76.0 
75.9 
75.9 
75.8 
75.7 
75.7 
75.6 
75.5 
75.5 
75.4 
75.3 


Cotain 


N.  sine.  X.  cos. 


10.5425041 
542027 ; 
541551  I 
541075  ! 
640600 
540125 
539651 
639177 
638703 
538230 
537758 

10.537286 
536814 
636342 
635871 
535401 
534931 
534461 
533992 
633524 
533055 

10.532587 
532120 
531653 
531186 
630720 
530254 
629789 
629324 
528859 
628395 

10.527932 
527468 
527005 
626543 
526081 
525619 
525158 
524697 
524237 
523777 

10.523317 
522858 
522399 
521941 
521483 
621025 
520568 
520111 
519655 
519199 

10.518743 
518288 
517833 
517379 
516925 
516471 
516018 
515565 
515113 
514661 


27564 
27592 
27620 
27648 
27676 
27704 
27731 
27759 
27787 
27815 
27843 
27871 
27899 
27927 
27955 
27988 
28011 
28039 
28067 
28095 
28123 
28150 
28178 
28206 
28234 
28262 
28290 
28318 
28346 
28374 
28402 
28429 
28457 
28485 
28513 
28541 
28569 
28597 
28625 
28652 
28680 
28708 
28736 
28764 
28792 
28820 
28847 
28875 
28903 
28931 
28959 
28987 
29015 
29042 
29070 
29098 
29126 
29154 
29182 
29209 
29247 


Tang.   Il  N.  cos.  N.sine.  ' 


96126 
96118 
96110 
96102 
96094 
96086 
96078 
96070 
96062 
96054 
96046 
96037 
96029 
96021 
96013 
96005 
y5997 
95989 
95981 
95972 
95964 
95956 
95948 
95940 
95931 
95923 
95915 
95907 
95898 
95890 
95882 
95874 
95865 
95857 
95849 
95841 
95832 
95824 
95816 
y5807 
95799 
95791 
95782 
95774 
95766 
95767 
95749 
95740 
95732 
95724 
95715 
95707 
95698 
95690 
95681 
95673 
95664 
95656 
95647 
95639 
95630 


73  Degrees. 


38 


Log.  Sines  and  Tangents.  (17°)  Natural  Sines. 


TABLE  II. 


Sine.  |D.  10" 


9 
10 
1! 
12 
13 
14 
16 

10 

17 
18 
19 

20 

21 

22 
23 
24 

25 
26 

27 
28 
29 

30 
31 
3-2 
33 
34 
86 
86 
37 
38 
39 
40 
41 
42 
43 
44 
45 
40 
47 
48 
40 
50 
51 
5-2 
53 
54 
55 
56 
57 
58 
59 
00 


9.465935 
406348 
466761 
467173 
467585 
467996 
468407 
468817 
469227 
469637 
470046 

9.470455 
470863 
471271 
471679 
472036 
472492 
472898 
473304 
473710 
474115 

9.474519 
474923 
475327 
475730 
476133 
476536 
476938 
477340 
477741 
478142 

9.478542 
478942 
479342 
479741 
480140 
480539 
480937 
481334 
481731 
482128 

9.482525 
482921 
483316 
483712 
484107 
484501 
484895 
485289 
485682 
486075 

9.486467 
486860 
487251 
487643 
488034 
488424 
488814 
489204 
489593 
489982 
Cosine. 


68.8 

68.8 

68.7 

68.6 

68.5 

68.5 

68.4 

68.3 

68.3 

68.2 

68.1 

68.0 

68.0 

67.9 

67.8 

67.8 

67.7 

67.6 

67.6 

67.5 

67.4 

67.4 

67.3 

67.2 

67.2 

67.1 

67.0 

66.9 

66.9 

66.8 

66.7 

66.7 

66.6 

66.5 

66.5 

66.4 

66.3 

66.3 

66.2 

66.1 

66.1 

66.0 

65.9 

65.9 

65.8 

65.7 

65.7 

65.6 

65.5 

65.5 

65.4 

66.3 

65.3 

65.2 

65.1 

65.1 

65.0 

65.0 

64.9 

64.8 


Cosine. 


9.980596 
980558 
980519 
980480 
980442 
980403 
980364 
980325 
980288 
980247 
980208 

). 980169- 
980130 
980091 
980052 
980012 
979973 
979934 
979895 
979855 
979816 

>. 979776 
979737 
979697 
979658 
979618 
979579 
979539 
979499 
979459 
979420 
9.979380 
979340 
979300 
979260 
979220 
979180 
979140 
979100 
979059 
979019 
9.978979 
978939 
978898 
978858 
978817 
978777 
978736 
978696 
978655 
978615  | 
978574 
978533 
978493 
978452 
978411 
978370 
978329 
978288 
978247 
978208 
"sTneT 


D.  10" 


6.4 

6.4 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.5 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.6 

6.7 

6.7 

6.7 

6.7 

6.7 

6.7 

6.7 

6.7 


Tain 


6.7 
6.7 
6.7 

6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 
6.8 


9.485339 
485791 
486242 
486693 
487143 
487593 
488043 
488492 
488941 
489390 
489838 

9.490286 
490733 
491180 
491627 
492073 
492519 
492965 
493410 
493854 
494299 

9.494743 
495186 
495630 
496073 
496515 
496957 
497399 
497841 
468282 
498722 

9.499163 
499603 
500042 
500481 
500920 
501359 
501797 
602235 
502672 
503109 

9.503546 
503982 
504418 
504854 
505289 
505724 
508159 
506593 
507027 
507460 

'.507893 
603326 
508759 
509191 
609622 
510054 
510485 
610916 
511346 
511776 
Cotang. 


D.  10' 


75.3 

75.2 

75.1 

75.1 

75.0 

74.9 

74.9 

74.8 

74.7 

74.7 

74.6 

74.6 

74.5 

74.4 

74.4 

74.3 

74.3 

74.2 

74.1 

74.0 

74.0 

74.0 

73.9 

73.8 

73.7 

73.7 

73.6 

73.6 

73.5 

73.4 

73.4 

73.3 

73.3 

73.2 

73.1 

73.1 

73.0 

73.0 

72.9 

72.8 

72.8 

72.7 

72.7 

72.6 

72.5 

72.5 

72.4 

72.4 

72.3 

72.2 

72.2 

72.1 

72.1 

72.0 

71.9 

71.9 

71.8 

71.8 

71.7 

71.6 


Cotanu-.  i  N.  sine. IN.  cos. 


10.514661  J I  29237 
514209  129265 
613758 I  29293 
513307  jj  29321 
512857! '29348 
612407 |  29376 
611957;;  29404 
511508  129432 
5110591!  29460 
510610  |29487 


95622 
95613 
95605 
95596 
95588 
95579 
95571 
95562 
95554 
5101621129515195645 


10.509714  129543 


609267 
608820 
508373 
507927 
507481 
507035 
503590 
508146 
605701 

10.505257 
504814 
504370 
503927 
503485 
503043 
502601 
502159 
601718 
501278 

10.500837 
600397 
499958 
499519 
499080 
498641 
498203 
497765 
497328 
496891 

10.496454 
496018 
495582 
495146 
494711 
494276 
493841 
493407 
492973 
492540!! 

10.492107 
491674 
491241 
490809 
490378 
489946 
489515 
489084 
488654 
488224 


! 29571 
! 29599 
| 29626 
! 29654 
i 29682 
129710 
| 29737 
I  29765 
1 29793 
29821 


95630 I 60 


29849-95441 


| 29876 
j 29904 
j 29932 
! 29960 
12998/ 
(30015 
j  30043 
1300/1 
30098 
I  30126 
130154 
130182 
I 30209 
30237 
30265 
30292 
30320 
30348 
30376 
30403 
30431 
30459 
30486 
30514 
30542 
30570 
3059/ 
30625 
30653 
30680 
30708 
30736 
3076  a 
30791 
30819 
30846 
30374 
30902 


95536 
95528 
95519 
95511 
95502 
95493 
95485 
95476 
95467 
95459 
95450 


95433 
95424 
95415 
95407 
95398 
95389 
95380 
95372 
95363 
95354 
95345 
95337 
95328 
95319 
95310 
95301 
95293 
95284 
95275 
95266 
95257 
95248 
95240 
95231 
95222 
95213 
95204 
95195 
95186 
95177 
95168 
95159 
95150 
95142 
95133 
95124 
95115 
95106 


Tan<: 


N.  cos.  N.si 


59 
58 
57 
56 
55 
54 
53 
52 
61 
50 
49 
48 
47 
48 
45 
44 
43 
42 
41 
40 
39 
38 
87 
36 
35 
34 
33 
82 
81 
30 
29 
28 
27 
26 
26 
•24 
23 
22 
21 
20 
19 
18 
17 
16 
16 
14 
13 
12 
11 
10 
9 
8 
7 
6 
5 
4 
8 
2 
1 
0 


79  Degrees. 


TABLE  IT. 


jog.  Sines  and  Tangents.    (18°)    Natural  Sines. 


39 


JSllK'. 

9.489982 
490371 
490759 
491147 
491535 
491922 
492308 
492695 
493081 
493466 
493851 

9.494236 
494621 
495005 
495388 
495772 
496154 
496537 
496919 
497301 
497682 

9.498064 
498444 
498825 
499204 
499584 
499963 
500342 
500721 
501099 
501476 

9.501854 
502231 
502607 
502984 
503360 
503735 
504110 
504485 
504860 
505234 

9.505608 
505981 
506354 
506727 
507099 
507471 
507843 
508214 
508585 
508956 

9.509326 
509696 
510065 
610434 
510803 
511172 
511540 
511907 
512275 
512642 


D.  10' 


Cosine. 


64.8 
64.8 
64.7 
64.6 
64.6 
64.5 
64.4 
64.4 
64.3 
64.2 
64.2 
64.1 
64.1 
64.0 
63.9 
63.9 
63.8 
63.7 
63.7 
63.6 
63.6 
63.5 
63.4 
63.4 
63.3 
63.2 
63.2 
63.1 
63.1 
63.0 
62.9 
62.9 
62.8 
62.8 
62.7 
62.6 
62.6 
62.5 
62.5 
62.4 
62.3 
62.3 
62.2 
62.2 
62.1 
62.0 
62.0 
61.9 
61.9 
61.8 
61.8 
61.7 
61.6 
61.6 
61.5 
61.5 
61.4 
61.3 
61.3 
61.2 


Cosine. 

.978206 
978165 
978124 
978083 
978042 
978001 
977959 
977918 
977877 
977835 
977794 
.977752 
977711 
977669 
977628 
977586 
977544 
977503 
977461 
977419 
977377 

9.977335 
977293 
977251 
977209 
977167 
977125 
977083 
977041 
976999 
976957 

9.976914 
976872 
976830 
976787 
976745 
976702 
976660 
976617 
976574 
976532 
976489 
976446 
976404 
976361 
976318 
976275 
976232 
976189 
976146 
976103 
976060 
976017 
975974 
975930 
975887 
975844 
975800 
975757 
975714 
975670 


D.  10" 


Sine. 


6.8 
6.8 
6.8 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
6.9 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.0 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.1 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 
7.2 


Tang. 

9.511776 
512206 
512635 
513064 
513493 
513921 
514349 
514777 
515204 
515631 
516057 

9.516484 
516910 
517335 
517761 
518185 
518610 
519034 
519458 
519882  I 
520305 
520728 
521151 
621573 
621995 
522417 
522838 
523259 
523680 
624100 
624520 
524939 
525359 
525778 
526197 
626615 
627033 
527451 
627868 
528285 
528702 
629119 
529535 
529950 
530366 
530781 
531196 
531611 
532025 
532439 
532853 
533266 
533679 
534092 
534504 
534916 
635328 
535739 
536150 
536561 
636972 


D.  10" 


Cotang. 


Cotang.  IN.sine.iN. 


10 


10-47927S 


10 


10 


10 


488224 ! 
487794  I 
487365  j 
486936  : 
486507  ! 
486079 ! 
485651 | 
485223  | 
484796 
484369 j 
483943  I 
483516  | 
483090  j 
482665  I 
482239  I 
481815 j 
481390 | 
480966 
480542  J 
480118 
479695 
479272 
478849 
478427 
478005 
47?583 
477162 
476741 
476320 
475900 
475480 
.476061 
474641 
474222 
473803 
473385 
472967 
472549 
472132 
471715 
471298 
.470881 
470465 
470050 
469634 
469219 
468804 
468389 
467975 
467561 
467147 
.466734 
466321 
465908 
465496 
465084 
464672 
464261 
463850 
463439 
463028 


Tang. 


30929 
30957 
30985 
31012 
31040 
31068 
31095 
31123 
31151 
31178 
31206 
31233 
31261 
31289 
31316 
31344 


31399 
31427 
31454 
31482 
31510 
31537 
31565 
31593 
31620 
31648 
31675 
31703 


95097 
95088 
95079 
95070 
95061 
95052 
95043 
95033 
95024 
95015 
95006 
94997 
94988 
94979 
94970 
94961 


31372i94952 


94943 
94933 
94924 
94915 
94906 
94897 
94888 
94878 
94869 
94860 
94851 
94842 


31730;94832 
31758194823 
31786194814 
31813J94805 
31841194795 
31868|94786 
31896194777 
31923194768 
3 1951 194758 
31979|94749 
32006194740 
32034;94730 
32061 194721 
3208994712 
32116194702 
32144(94693 
32171194684 
32199J94674 
32227  94665 
32250!94656 
32282J94646 
3230994637 


32337 
32364 
32392 
32419 
32447 
32474 
32502 
32529 
32557 


94627 
94618 
94609 
94599 
94590 
94580 
94571 
94561 
94552 
N.  cos.jN.sine, 


71  Degrees. 


43 


Log.  Sines  and  Tangents.  (19°)  Natural  Sines. 


TABLE  II. 


0 

9.512042 

1 

5i300y 

2 

513375 

3 

513741 

4 

514107 

5 

514472 

6 

514837 

7 

515202 

8 

515500 

9 

515930 

10 

510294 

11 

9.510057 

12 

517020 

13 

517382 

14 

517745 

15 

518107 

16 

518468 

17 

518829 

18 

519190 

19 

519551 

20 

519011 

21 

9.520271 

22 

520031 

23 

520990 

24 

521349 

25 

521707 

26 

522000 

27 

522424 

28 

522781 

29 

523138 

31) 

523495 

31 

9.523852 

32 

524208 

33 

524504 

31 

524920 

35 

525275 

36 

525680 

37 

525984 

38 

526339 

39 

520093 

40 

527040 

41 

9.527400 

42 

527753 

43 

528105 

44 

528458 

45 

528810 

46 

529101 

47 

529513 

48 

529804 

4<j 

530215 

60 

530505 

51 

3.530915 

52 

531205 

53 

531014 

54 

531903 

55 

532^12 

86 

532001 

57 

533009 

58 

53335  7 

50 

533  704 

60 

534052 

U.  10' 

01.2 
61.1 
01.1 
61.0 
60.9 
00.9 
60.8 
60.8 
60.7 
60.7 
60.6 
60.5 
60.5 
60.4 
60.4 
60.3 
00-3 
60-2 
60.1 
60.1 
60.0 
60.0 
59.9 
59.9 
59.8 
59.8 
59.7 
59.6 
59.6 
59.5 
59.5 
59.4 
59.4 
59.3 
59.3 
59.2 
59.1 
59-1 
59-0 
59-0 
58-9 
58-9 
58.8 
58.8 


Cosine. 


58 

58 

58 

58 

58 

58 

58 

58-4 

58-3 

58-2 

58-2 

58.1 

58.1 

58.0 

58.0 

57.9 


9.9 


.975070 
976627 
975583 
9  75539 
975490 
975452 
975408 
975305 
975321 
975277 
975233 
.975189 
975145 
975101 
975057 
975013 
974909 
974925 
974880 
974836 
974792 
4748 
974703 
974659 
974014 
^74570 
974525 
974481 
974436 
974391 
974347 
974302 
974257 
974212 
974167 
974122 
974077 
974032 
973987 
973942 
973897 
973852 
973807 
973761 
973716 
973671 
973625 
973580 
973535 
973489 
973444 
973398 
973352 
973307 
973201 
973215 
973169 
973124 
973078 
973032 
972986 


9.97 


Sine. 


7.3 

7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.3 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.4 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
7.5 
5 
5 
6 
(i 


7.6 
7.6 
7.6 
7.6 
7.6 
7.6 
7.6 
7.6 
7.6 
7.6 
7.7 


9.530972 
•  537382 
537792 
538202 
538011 
539020 
539429 
539837 
540245 
540053 
541061 
9.541468 
541875 
542281 
542088 
543094 
543499 
543905 
544310 
544715 
545119 
545524 
545928 
546331 
546735 
547138 
547540 
547943 
548345 
548747 
549149 
9.549550 
549951 
550352 
650752 
651152 
551552 
551952 
552351  I 
552750 
553149 
9.553548 
553946 
554344 
554741 
555139 
555536 
555933 
556329 
556725 
557121 
9.557517 
557913 
558308 
558702 
659097 
559491 
559885 
560279 
560673 
561006 
Cotang. 


l>.  iu  J  Cotang.  ||N.  sine.|N.  cos.| 


68.4 
68.3 
68.3 
68.2 
68.2 
68.1 
68.1 
68.0 
68.0 
67.9 
67.9 
67.8 
67.8 
67.7 
67.7 
67.6 
67.6 
67.5 
67.5 
67.4 
67.4 
67.3 
67.3 
67.2 
67.2 
67.1 
67.1 
67.0 
67.0 
66.9 
66.9 
66.8 
66.8 
66.7 
66.7 
66.6 
66.6 
5 


10.463028  3255 


66, 

00. 

00. 

00. 

GO. 

66. 

00. 

66. 

66.2 

66.1 

66.1 

66.0 

66.0 

65.9 

65.9 

65.9 

65.8 

65.8 

65.7 

65.7 

65.6 

65.6 

65.5 


402018  :i258; 
402208 

401798  32039 
401389  32667 
400980 i  32694 
400571 I  32722 
400163  I  32749 
459755  32777 
459347;  32804 
458939:  32832 

10.458532  32859 
458125  I  32887 
457719  132914 
457312  32942 
450900  132909 
450501  32997 
456095  : 33024 
455690  133051 
455285!  33079 
454881  33100 

10.454476  33134 
454072  j 33161 
453669;  133189 
4532651:33216 
452862  !j  33244 
452400  j 33271 
452057  !'  33298 
451055  33320 
451253  33353 
450851  i|  33381 

10.450450  ! 33408 
450049  133436 
449648  33463 
449248  33490 
448848  33518 
418448  ' 33545 
448048  33573 
447049  33000 
447250  133627 
446851 

10.446452 

440054  33710J94147 


94552 
94542 

94533 
94523 
94514 
94504 
94495 
94485 
94476 
94466 
94457 
94447 
94438 
94428 
94418 
94409 
94399 
94390 
94380 
94370 
94361 
94351 
94342 
94332 
94322 
94313 
94303 
94293 
94284 
94274 
94264 
94254 
94245 
94235 
94225 
94215 
94206 
94196 
94186 
94176 


33655  94167 
33682  94157 


445656  33737 
445259  33764 
444861  33  792 
444404  33819 
444067 '133846 


443671 
443275 
442879 
10.442483 
442087 
441692 
441298 
440903 
440509 


33874 
33901 
33929 


33950  94058 


33983 
34011 
34038 
34065 
34093 


440115  34120 


439721  : 
439327  34175 
438934  34202 


Tang. 


N.  cos.  X.sine 


94137 
94127 
94118 
94108 
94098 
94088 
94078 
94068 


94049 
94039 
94029 
94019 
94009 
93999 


34147  93989 
93979 
93969 


00 
59 

58 
57 
^ 
66 
54 
53 
52 
51 
50 
49 
48 
47 
40 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
85 
84 
83 
82 
81 
30 
20 

28 

27 

20 

25 

24 

23 

22 

21 

•JO 

10 

18 

17 

10 

15 

14 

13 

12 

11 

10 
0 
8 
7 
6 
5 
4 
3 
2 
1 
0 


70  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (2C°)    Natural  Sines. 


11 


534399 

534745 
53509-2 
635438 
535783 
536129 
536474 
536818 
537163 
53750/ 

9.537851 
538194 
538538 
538880 
539223 
539565 
539907 
540249 
540590 
540931 

9.541272 
541613 
541953 
542293 
542632 
542971 
543310 
543649 
543987 
544325 

9.544663 
545000 
545338 
545674 
546011 
546347 
546683 
547019 
547354 
547689 

9.548024 
548359 
548693 
549027 
549360 
549693 
550026 
550359 
550692 
551024 

9.551356 
551687 
552018 
552349 
552680 
553010 
553341 
553(570 
554000 
554329 


Cosine. 


D.  10"  Cosine.  D.  lu 


57.8 

57.7 

57.7 

57-7 

57 

57 

5  V 

57 

57 

57.3 

57.3 

57.2 

57.2 

57.1 

57.1 

57.0 

57.0 

56.9 

56.9 

56.8 

56.8 

56.7 

56.7 

56.6 

56.6 

56.5 

56.5 

56.4 

56.4 

56.3 

56.3 

56.2 

56.2 

56.1 

56.1 

56.0 

56.0 

55.9 

55.9 

55.8 

55.8 

55.7 

55.7 

55.6 

55.6 

55.5 

55.5 

55.4 

55.4 

55.3 

55.3 

55.2 

55.2 

55.2 

55.1 

55.1 

55.0 

55.0 

54.9 

54.9 


9.972986 

972940 
972894 
972848 
972802 
972755 
972 70 J 
9J 2663 
972617 
972570 
972524 

9.972478 
972431 
9*2385 
972338 
972291 
972245 
972198 
972151 
972105 
972058 

9.972011 
971964 
971917 
971870 
971823 
971776 
971729 
971682 
971635 
971588 

9.971540 
971493 
971446 
971398 
971351 
971303 
971256 
971208 
971161 
971113 

9.971066 
971018 
970970 
970922 
970874 
970827 
970779 
970731 
970683 
970635 

9.970586 
970538 
970490 
970442 
970394 
970345 
970297 
970249 
970200 
970152 


Sine. 


7.7 

7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.8 
7.9 
7.9 
7.9 
7.9 
7.9 


7.9 
7.9 
7.9 
7.9 
7.9 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.0 
8.1 
8.1 
8.1 
8.1 


Tang. 


9.561066 
561459 
561851 
562244 
562636 
563028 
563419 
563811 
564202 
564592 
564983 

9.565373 
565763 
566153 
566542 
566932 
667320 
567709 
568098 
568486 
568873 

9.569261 
569648 
570035 
570422 
570809 
571195 
571581 
571967 
572352 
572738 

9.573123 
573507 
573892 
574276 
574660 
575044 
575427 
575810 
676193 
576576 

9.576958 
577341 
577723 
578104 
578486 
578867 
579248 
579629 
580009 
580389 

9.580769 
581149 
581528 
581907 
582286 
582665 
683043 
583422 
583800 
684177 


Cotam 


D.  10 


65.5 
65.4 
65.4 
65  3 
65.3 
65.3 
65.2 
65.2 
65.1 
65.1 
65.0 
65.0 
64.9 
64.9 
64.9 
64  8 
64.8 
64.7 
64.7 
64.6 
64.6 
64.5 
64.5 
64.5 
64.4 
64.4 
64.3 
64.3 
64.2 
64.2 
64.2 
64.1 
64.1 
64.0 
64.0 
63.9 
63.9 
63.9 
63.8 
63.8 
63.7 
63.7 
63.6 
63.6 
63.6 
63.5 
63.5 
63.4 
63.4 
63.4 
63.3 
63.3 
63.2 
63.2 
63.2 
63.1 
63.1 
63.0 
63.0 
62.9 


Co  tang. 

10.438934 
438541 
438149 
437756 
437364 
436972 
436581 
436189 
435798 
435408 
435017 

10.434627 
434237 
433847 
433458 
433068 
432680 
432291 
431902 
431514 
431127 

10.430739 
430352 
429965 
429578 
42919*1 
428805 
428419 
428033 
427648 
427262 

10.426877 
426493 
426108 
425724 
425340 
424956 
424573 
424190 
423807 
423424 

10.423041 
422659 
422277 
421896 
421514 
421133 
420752 
420371 
419991 
419611 

10.419231 
418851 
418472 
418093 
417714 
417335 
416957 
416578 
416200 
415823 

Tang.   I 


N.  sine. 

1 34202 
! 34229 
! 34257 
! 34284 
3431 
! 34339 
i  34366 
! 34393 
i 34421 
! 34448 
34475 
34503 
34530 
3455 
34584 
34612 
34639 
34666 
34694 
34721 
34748 
34775 
34803 
34830 
34857 
34884 
34912 
34939 
34966 
34993 
35021 
35048 
35075 
35102 
35130 
35157 
35184 
35211 
35239 
35266 
35293 
35320 
35347 
35375 
35402 
35429 
35456 
35484 
35511 
35538 
35565 
35592 
35619 
35647 
35674 
35701 
35728 
35755 
35782 
35810 
35837 


N.coa. 

93969 
93959 
93949 
93939 
93929 
93919 
93909 
93899 
93889 
93879 
93869 
93859 
93849 
93839 
93829 
93819 
93809 
93799 
93789 
93779 
93769 
93759 
93748 
93738 
93728 
93718 
93708 
93698 
93688 
93677 
93667 
93657 
93647 
93637 
93626 
93616 
93606 
93596 
93585 
93575 
93565 
93555 
93544 
93534 
93524 
93514 
93503 
93493 
93483 
93472 
93462 
93452 
93441 
93431 
93420 
93410 
93400 
93389 
93379 
93368 
93358 

N.  cos.  N.sine. 


69  Degrees. 


4:2 


Log.  Sines  and  Tangents.     (21°)    Natural  Sines. 


TABLE  II. 


0 
1 
2 
3 
4 
5 
6 
7 
8 
9 

10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
•20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 


Sine. 

3.554329 
554658 
554987 
555315 
555643 
555971 
556299 
556626 
556953 
557280 
557608 

9.557932 
558258 
558583 
558909 
559234 
55y558 
559883 
560207 
560531 
560855 

9.561178 
561501 
561824 
562146 
582468 
562790 
563112 
563433 
563755 
564075 

9.564396 
564716 
565036 
565356 
565676 
565995 
566314 
566632 
566951 
567269 

9.567587 
567904 
568222 
568539 
568856 
569172 
569488 
569804 
570120 
570435 

9.570751 
571066 
571380 
571695 
572009 
572323 
572636 
572950 
573263 
573575 
Cosine. 


D.  10"[  Cosine. 


54.8 
54.8 
54.7 
54.7 
54.6 
54.6 
54.5 
54.5 
54.4 
54.4 
54.3 
54.3 
54.3 
54.2 
54.2 
54.1 
54.1 
54.0 
54.0 
53.9 
53.9 
53.8 
53.8 
53.7 
53.7- 
53.6 
53.6 
53.6 
53.5 
53.5 
53.4 
53.4 


53 

53 

53 

53 

53 

53.1 

53.1 

53.0 

53.0 

52.9 

52.9 

52.8 

52.8 

52.8 

52.7 

52.7 

52.6 

52.6 

52.5 

52.5 

52.4 

52.4 

52.3 

52.3 

52.3 

52.2 

52.2 

52.1 


9.970152 
970103 
970055 
970006 
969957 
969909 
939860 
969811 
989762 
969714 
969665 
3.969616 
969567 
969518 
969469 
969420 
969370 
969321 
969272 
969223 
989173 
3.969124 
969075 
969025 
968976 
968926 
968877 
968827 
968777 
968728 
968678 
>.  968628 
968578 
968528 
968479 
968429 
968379 
988329 
968278 
968228 
968178 
9.988128 
9680 ?8 
968027 
967977 
967927 
967876 
937826 
967775 
967725 
967674 
9.967624 
987573 
957522 
967471 
987421 
967370 
967319 
987268 
987217 
96,166 


D.  10"i   Tang. 


8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.1 
8.2 
8.2 


8 

S 

8 

8 

8 

8 

8 

8 

8 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.3 

8.4 

8.4 

8.4 

8.4 

8.4 

8.4 

8.4 

8.4 

8.4 

8,4 

8.4 

8.4 

8.4 

8.4 

8.5 

8.5 

8.5 

8.5 

8.5 

8.5 

8.5 


9.584177 
584555 
584932 
585309 
585888 
588062 
586439 
586815 
587190 
587566 
587941 
9.588316 
588891 
589066 
589440 
589814 
590188 
590562 
590935 
591308 
591681 
9.592054 
592426 
592798 
693170 
593542 
593914 
594286 
594656 
595027 
595398 
9.595768 
596138 
596508 
596878 
597247 
597616 
597985 
598354 
598722 
599091 
599459 
599827 
600194 
600562 
600929 
601296 
601662 
602029 
602395 
602761 
9.603127 
603493 
603858 
604223 
604588 
604953 
605317 
605682 
(  08046 
608410 


0.  10' 


Sine. 


Cotang. 


62.9 

62.9 

62.8 

62.8 

62.7 

62.7 

62.7 

62.6 

62.6 

62.5 

62.5 

62.5 

62.4 

62.4 

62.3 

62.3 

62.3 

62.2 

62.2 

62.2 

62.1 

62.1 

62.0 

62.0 

61.9 

61.9 

61.8 

61.8 

61.8 

61.7 

61.7 

61.7 

61.6 

61.6 

61.6 

61.5 

61.5 

61.6 

61.4 

61.4 

61.3 

61.3 

61.3 

61.2 

61.2 

61.1 

61.1 

61.1 

61.0 

61.0 

61.0 

60.9 

60.9 

60.9 

60.8 

60.8 

60.7 

60.7 

60.7 

60.6 


Cotam 


10.415823 
415445 
415038 
414691 
414314 
413938 
413561 
413185 
412810 
412434 
412059 

10.411684 
411309 
410934 
410560 
410186 
409812 
409438 
409065 
408692 
408319 

10.407946 
407574 
407202 
406829 
406458 
406086 
405715 
405344 
404973 
404602 

10.404232 
403862 
403492 
403122 
402753 
402384 
402015 
401646 
401278 
400909 

10.400541 
400173 
399806 ! 
399438 ! 
399071 | 
398704 ' 
398338 ' 
397971 
397605 
397239  J 
10.396873  | 
396507|j 
396142!| 
395777! j 
395412  ; 
395047 
394683 
394318 !  i 
393954  ' 
393590  :| 


35837 
35864 


X.  cos. 


93358 
93348 
35891  93337 


35918 
35945 
3597^ 
36000 
36027 
36054 
36081 
36108 
36135 
36162 
36190 


36217  93211 


36244 
36271 
36298 
36325 


93327 
93316 
93308 
93295 
93285 
93274 
93264 
93253 
93243 
93232 
93222 


93201 

93190 

93180 

3169 


36352  93159 


36379 
36406 
36434 
36461 
36488 
36515 
36542 
36569 
3659t 
36623 
36650 
3667 
36704 
36731 
36758 
36785 
36812 
36839 
3686 
! 36894 
36921 
1 36948 
! 36975 
| 37002 
J37029 


37056  92881 


1 37083 
37110 
i 37137 
37164 
137191 
37218 
37245 
37272 
37299 
37326 
37353 
37380 
37407 
37434 
37461 


93148 
93137 
93127 
y3116 
93108  I  36 
93095  I  35 


93084 
93074 
93063 
93052 
93042 
93031 
93020 
93010 
92999 
92988 
92978 
92967 
92956 
92945 
92935 
92926 
92913 
92902 
92892 


92870 
92859 
92849 
92838 
32827 
92816 
92805 
92794 
92784 
92773 
92762 
92751 
2740 
92729 
92718 


Tang. 


IN.  cos.  A. sine 


68  Decrees 


TABLE  II. 


Log.  Sines  and  Tangents.     (22°)     Natural  Sines. 


43 


D.  10"  Cosi 


.573575 
573888 
5  74200: 
574512 
574824 
575136 
575447 
575/58 
576059 
570379 
570089 

.576999 
577309 
577018 
577927 
578230 
578545 
578853 
579102 
579470 
579777 

.580035 
580392 
580699 
581005 
581312 
581618 
681924 


582229  fc.,, 
582535  ~' 


582840 
.583145 
583449 
583754 
584058 
584301 
584065 
584968 
585272 
585574 
585877 
.586179 
586482 
586783 
587085 
587386 
587688 
587989 
588289 
588590 
588890 
.589190 
589489 
589789 
590088 
590387 
590686 
590984 
591282 
591580 
591878 
Cosine. 


,967166 
967115 
967064 
907013 
966961 
966910 
906859 
96880S 
906756 
966705 
966653 
.966602 
966550 
966499 
966447 
966395 
966344 
966292 
966240 
966188 
966136 

9.966085 
966033 
965981 
965928 
965876 
965824 
965772 
965720 
965668 
965615 

9.965563 
965511 
965458 
965403 
965353 
965301 
965248 
965195 
965143 
965090 
965037 
964984 
964931 
964879 
964826 
964773 
964719 
964666 
964613 
964560 
.964507 
964454 
964400 
964347 
964294 
964240 
964187 
964133 
964080 
9J4026 


Sme. 


D.  10' 


8.5 
8.5 
8.5 
8.5 
8.5 
8.5 
8.5 
8.5 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.6 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.7 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.8 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 
8.9 


Tans 


606410 
006773 
607137 
607500 
607863 
608225 
608588 
008950 
609312 
609674 
610030 
010397 
610759 
611120 
611480 
611841 
612201 
612561 
612921 
613281 
613641 
614000 
614359 
614718 
615077 
615435 
615793 
616151 
616509 
616867 
617224 
9.617582 
617939 
618295 
618652 
619008 
619364 
619721 
620076 
620432 
620787 
621142 
621497 
621852 
622207 
622561 
622915 
623269 
623623 
623976 
624330 
624683 
625036 
625388 
625741 
626093 
626445 
626797 
627149 
627501 
627852 


Cotang. 


D.  10" 


60.6 
60.6 
60.5 
60.5 
60.4 
60.4 
00.4 
60.3 
60.3 
60.3 
60.2 
60.2 
60.2 
60.1 
60.1 
60.1 
60.0 
60.0 
60.0 
59.9 
59.9 
59.8 
59.8 
59.8 
59.7 
59.7 
59.7 
59.6 
59.6 
59.6 
59.5 
59.5 
59.5 
59.4 
59.4 
59.4 
59.3 
59.3 
59.3 
59.2 
59.2 
59.2 
59.1 
59.1 
59.0 
59.0 
59.0 
58.9 
58.9 
58.9 
58.8 
58.8 
58.8 
58.7 
58.7 
58.7 
58.6 
58.6 
58.6 
58.5 


Cotang.  j  N .  sine.l N.  cos . 


10.393590 
S93227 
392863 
392500 
392137 
391775 
391412 
39105t) 
390888 
390326 
389964 

10.389603 
389241 
388880 
388520 
388159 
387799 
387439 
387079 
386719 
386359 

10-386000 
385641 
385282 
384923 
384565 
384207 
383849 
383491 
383133 
382776 

10-382418 
382061 
381705 
381348 
380992 
380636 
380279 
379924 
379568 
379213 

10-378858 
378503 
378148 
377793 
377439 
377085 
376731 
376377 
376024 
375670 

10-375317 
374964 
374612 
374259 
373907 
373555 
373203 
372851 
372499 
372148 


37461192718 


!  37488 

!  37515 

:;  37542 

!  37569 

1137595 

1 1 37622 

37649 

37676 

37703 

37730 

37757 

37784 

37811 


92707 
92697 
92686 
92675 
92664 
92653 
92642 
92631 
92620 
92609 
92598 
92587 
92576 


37838  92565 


37865 
37892 
37919 
37946 
37973 
37999 
38026 
38053 
38080 
38107 
38134 
38161 
38188 
38215 
38241 
38268 
38295 
38322 
38349 
38376 
38403 


92554 
92543 
92532 
92521 
92510 
92499 
92488 
92477 
92466 
92455 
92444 
92432 
92421 
92410 
92399 
92388 
92377 
92366 
92355 
92343 
92332 


3843092321 


38456 
38483 
38510 
38537 
38564 
3.8591 
38617 
38644 
38671 
38698 
38725 
38752 
38778 
38805 
38832 
38859 
38886|92130 
38912  92119 
3893992107 
J92096 


92310 
92299 
92287 
92276 
92265 
92254 
92243 
92231 
92220 
92209 
92198 
92186 
92175 
92164 
92152 
92141 


38966 
38993 
39020 
39046 


92085 
92073 
92062 


39073|92050 


Tang.   |  N.  cos.)  N.sine, 


67  Degrees. 


44 


Log.  Sines  and  Tangents.     (23°)    Natural  Sines. 


TABLE  II. 


0 
1 

2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
1? 
IS 
1!) 
20 
21 
22 
23 
24 
25 
25 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
40 
50 
51 
52 
53 
54 
55 
56 
57 
58 
50 
60 


9.591878 
592176 
592473 
592770 
593087 
593363 
593659 
593955 
594251 
594547 
594842 

9.595137 
595432 
595727 
595021 
596315 
598609 
596903 
597196 
597490 
597783 
598075 
598368 
598660 
598952 
599244 
599536 
599827 
600118 
600409 
600700 
600990 
601280 
601570 
601860 
602150 
602439 
602728 
603017 
603305 
603594 

9.603882 
604170 
604457 
604745 
605032 
605319 
605606 
605892 
603179 
603465 
.608751 
607036 
607322 
607607 
607892 
608177 
608461 
608745 
609029 
609313 
Cosine,  i 


D.  10' 


49.8 
49.5 
49.5 
49.5 
49.4 
49.4 
49.3 
49.3 
49.3 
49.2 
49.2 
49.1 
49.1 
49.1 
49.0 
49.0 
48.9 
48.9 
48.9 
48.8 
48.8 
'48.7 
48.7 
48.7 
48.6 
48.6 
48.5 
48.5 
48.5 
48.4 
48.4 
48.4 
48.3 
48.3 
48.2 
48.2 
48.2 
48.1 
48.1 
48.1 
48.0 
48.0 
47.9 
47.9 
47.9 
47  8 
47.8 
47.8 
47.7 
47.7 
47-6 
47-6 
47-6 
47-5 
47-5 
47-4 
47.4 
47.4 
47.3 
47.3 


Cosine. 

1.964026 
963972 
963919 
963865 
963811 
963757 
983704 
963650 
963596 
963542 
963488 
'.963434 
963379 
963325 
963271 
963217 
963163 
963108 
963054 
962999 
962945 
.962890 
962836 
962781 
962727 
962672 
962617 
962562 
962508 
962453 
962398 
.962343 
962288 
962233 
962178 
962123 
962067 
962012 
961957 
961902 
961846 
.961791 
961735 
961680 
961624 
961569 
961513 
961458 
961402 
961346 
961290 
.961235 
961179 
961123 
961087 
961011 
960955 
960899 
960843 
960786 
930730 
Sine. 


8.9 

8.9 

8.9 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.0 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9.1 

9  2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.2 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.3 

9.4 

9.4 

T;ti!<j 


9.627852 
628203 
628554 
628905 
629255 
629606 
629956 
630308 
630656 
631005 
631355 
9.631704 
632053 
632401 
632750 
633098 
633447 
633795 
634143 
634490 
634838 
9.635185 
635532 
635879 
636226 
636572 
636919 
637265 
637611 
637956 
638302 
.638647 
638992 
639337 
639682 
640027 
640371 
640716 
641060 
641404 
641747 
,642091 
642434 
642777 
643120 
643463 
643806 
644148 
644490 
644832 
(,45174 
645516 
645857 
646199 
646540 
646881 
647222 
647562 
647903 
648243 
648583 


Cotang. 


58 

58 

58 

58 

58 

58 

58 

58.2 

58.1 

58.1 

58.1 

58.0 

58.0 

58.0 

57.9 

57.9 

57.9 

57.8 

57.8 

57.8 

67.7 

57.7 

57.7 

57.7 

57.6 

67.6 

67.6 

57.5 

57.5 

57.5 

57.4 

57.4 

57.4 

57.3 

57.3 


Cotang.   ft!.,  sine.  N.  cos. 


57 

57 

57 

57 

57 

57 

57.1 

57.1 

57.0 

57.0 

57.0 

56.9 

56.9 

56.9 

56.9 

56.8 

56.8 

56.8 

56.7 

56.7 

58.7 


10.372148 
371797 
371446 
371095 
370745 
370394 
370044 
369694 
369344 
368995 
368645 
10.368296 
367947 
367599 
367250 
366902 
366553 
366205 
365857 
365510 
365162 
10.364815 
364468 
364121 
363774 
363428 
363081 
362735 
362389 
362044 
361698 
10.361353 
361008 
360663 
360318 
359973 
359629 
359284 
368940 
358596 
358253 
10.357909 
357566 
357223 
356880 
356537 
356194 
355852 
355510 
355168 
354826 
10.354484 
354143 
353801 
353460 
353119 
352778 
352438 
352097 
351757 
351417 


39073 
| 39100 
39127 
39153 
j  1 39180 
39207 
j:  39234 
!  139260 
39287 
1139314 
i | 39341 
i 139367 
39394 
39421 
39448 
39474 
39501 
39528 
39555 
39581 
39608 
39635 
39661 
39688 
39715 
39741 
39768 
39795 
3%22 
39848 
39875 
39902 
39928 
39955 
39982 
40008 
40035 
40062 
40088 
40115 
40141 
40168 
40195 
40221 
40248 
40275 
40301 
40328 
40355 
40381 
40408 
40434 
40461 
40488 
40514 
40541 
40567 
40594 
40621 
40647 
40674 


Tang. 


92050 

92039 

92028 

92016 

92005 

91994 

91982 

91971 

91959 

91948 

91936 

91925 

91914 

91902 

91891 

91879 

91868 

91856 

91845 

91833 

91822 

91810 

91799 

91787 

91775 

91764 

91752 

91741 

91729 

91718 

91706 

91694 

91683 

91671 

91660 

91648 

91636 

91625 

91613 

91601 

91590 

91578 

91566 

91555 

91543 

91531 

91519 

91508 

91496 

91484 

91472 

91461 

91449 

91437 

91425 

91414 

91402 

91390 

91378 

91366 

91355 


N.  cos.  N.sine. 


60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
36 
34 
33 
32 
31 
30 
29 
28 
27 
26 
25 
24 
23 
^2 
21 
20 
19 
18 
17 
16 
15 
N 
13 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0 


66  Degrees. 


TABLE  H. 


Log.  Sines  and  Tangents.    (24°)     Natural  Sines. 


45 


Sine. 

9.609313 
609597 
609880 
610164 
610447 
6107-29 
611012 
611294 
611576 
611858 
612140 

9.612421 
612702 
612983 
613264 
613545 
613825 
614105 
614385 
614665 
614944 

9.615223 
615502 
615781 
6160S0 
616338 
616616 
616894 
617172 
617450 
617727 

9.618004 
618281 
618558 
618834 
619110 
619386 
619662 
619938 
620213 
620488 

9.620763 
621038 
621313 
621587 
621861 
622135 
622409 
622682 
622956 
623229 
9.623512 
623774 
624047 
624319 
624591 
624863 
625135 
625403 
625677 
625948 


I).  10 


I  Cosh 


47.3 

47.2 

47.2 

47.2 

47.1 

47.1 

47.0 

47.0 

47.0 

46.9 

46.9 

46.9 

46.8 

46.8 

46.7 

46.7 

46.7 

46.6 

46.6 

46.6 

46.5 

46.5 

46.5 

46.4 

46.4 

46.4 

46.3 

46.3 

46.2 

46.2 

46.2 

46.1 

46.1 

46.1 

46.0 

46.0 

46.0 

45.9 

45.9 

45.9 

45.8 

45.8 

45.7 

45.7 

45.7 

45.6 

45.6 

45.6 

45.5 

45, 

45 

45, 

45 

46, 

45. 

45, 

46. 


Cosine. 


45.2 
45.2 


960730 
960674 
960618 
960561 
960505 
960448 
960392 
960335 
960279 
960222 
960165 
960109 
960052 
959995 
959938 
959882 
959825 
959768 
959711 
959654 
959596 

9.959539 
959482 
959425 
959368 
959310 
959253 
959195 
959138 
959081 
959023 
958965 
958908 
958850 
958792 
958734 
958677 
958619 
958561 
958503 
958445 

9.958387 
958329 
958271 
958213 
958154 
958096 
958038 
957979, 
957921 
957863 
957804 
957746 
957687 
957628 
957570 
957511 
957452 
957393 
957335 
957276 


Sine. 


9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.4 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.5 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.6 
9.7 


D.  It/   Tang.   D.  lu'  Cotang. 

648583  .R  R  10.351417 

648923  ?°'°  351077 

649263  ?X°  350737 

649602  gJj'U  350398 

649942  gj  r  350058 

650281  j£  •?  349719 

650620  JJ  349380 

650959  gj'°  349041 

651297  °X'7  348703 

651636  2M  348364 

651974  °2'q  348026 

652312  £°"q  10.347688 

652650  ™'t  347350 

652988  ?"•*  347012 

653326  2!*o  346674 

653663  ?°'o  346337 

654000  °°-*  346000 

654337  2?  i  345663 

654174  25 'J  345326 

655011  25-}  344989 

655348  2M  344652 

9.655684  ™*n  10.344316 

656020  2J'n  343980 

656356  25*  n  343644 

656692  2?'"  343308 

657028  2?'q  342972 

657364  2?q  342636 

657699  2!'q  342301 

658034  2?"e  341966 

658369  ?r'«  341631 

658704  °8'e  341296 

9.659039  2J"e  10.340961 

659373  2?'?  340627 

659708  98  •'  340292 

660042  ?;?•'  339958 

660376  ?r  7  339624 

660710  2?'«  339290 

661043  2?  «  338957 

661377  2^£  338623 

661710  *??•£  338290 

662043  2?'£  337957 

9.662376  £?•?  10.337624 

662709  2?  a  337291 

663042  2! 'I  336958 

663375  J??*?  336625 

663707  rS-7  336293 

664039  2? 'J  335961 

664371  2J'q  335629 

664703  2J'q  335297 

665035  °°',  334965 

665366  r?"o  334634 

). 665697  2?'o  10.334303 

666029  £2  9  333971 

666360  2? -7  333620 

666691  2!  1  333309 

667021  2^1  332979 

667352  °°'\  332648 

667682  ?)?•*  332318 

668013  J£n  331987 

668343  rg'JJ  331657 

668672  °°-U  331328 


9.7 
9.7 
9.7 
9.7 
9.7 
9.7 
9.7 
9.8 
9.8 
9.8 
9.8 
9.8 
9.8 
9.8 
9.8 


Cotang. 


3? 


40674191355 
4070091343 
40727  .91331 
40753  91319 
40780  91307 


40800 
40833 
40860 
40886 
140913 
40939 
I  40966 
I  40992 
141019 
41045 
41072 
! 41098 
41125 
41151 
41178 
41204 
41231 


41284 


41310  91068 
41337  91056 
41363  91044 
41390  91032 
41416  91020 
41443  91008 
41469  90996 
41496  90984 
41522 


91295 
91283 
91272 
91260 
91248 
91236 
91224 
91212 
91200 
91188 
91176 
91164 
91152 
91140 
91128 
91116 
91104 


41257  91092 


91080 


41549 
41575 


90972 
90960 

90948 


41602  90936 
41628  J90924 
41655  90911 
4168190899 
41707J90887 
41734190875 
41760  90863 


41787 
41813 
41840 
41866 
41892 


90851 
90839 
90826 
90814 
90802 


4191990790 

41945190778 

41972  90766 

41998 

42024 

42051 

42077 

42104 


42130 
42156 
42183 
42209 
42235 
42262 


90753 
90741 
90729 
90717 
90704 
90692 
90680 
90668 
90655 
90643 
90631 


Fang.        j  N.  eos.  .\.sinc 


60 
59 

68 
57 
56 
55 
64 
53 
52 
61 
50 
49 
48 
47 
if> 
45 
44 
43 
42 
41 
40 
39 

37 
06 
35 
34 
33 
32 
31 
30 
29 
•28 
27 
20 
25 
24 
23 
22 
21 

m 

19 
18 
17 
16 
15 
14 
13 
12 
11 
10 

9 

8 

7 

6 

5 

4 

3 

2 

1 

0 


65  Degrees. 


46 


Log.  Sines  and  Tangents.     (25°)     Natural  Fines. 


TABLE  II. 


Sine.      ID.  10" 


0 

9. 625948 

1 

626219 

2 

625490 

3 

636750 

4 

627030 

5 

627300 

6 

627570 

7 

627840 

8 

628109 

9 

6283 ?8 

If) 

628647 

11 

9.628913 

12 

629185 

13 

629453 

14 

629721 

15 

620989 

lii 

630257 

17 

630524 

18 

630792 

19 

631059 

20 

631326 

21 

9.631593 

22 

631859 

33 

632125 

2-i 

632392 

25 

632658 

36 

632923 

27 

633189 

38 

633454 

39 

633719 

30 

633984 

31 

9.634249 

32 

634514 

33 

634778 

34 

635042 

33 

635305 

36 

635570 

37 

635834 

38 

63609? 

39 

636360 

40 

636623 

•it 

9.635886 

43 

637148 

43 

637411 

44 

637673 

45 

637935 

46 

638197 

47 

638458 

48 

638720 

49 

638981 

60 

639242 

51 

9.639503 

5-2 

639764 

53 

640024 

54 

640284 

55 

640544 

56 

640304 

5? 

641064 

58 

641324 

59 

641584 

60 

641842 

Cosine. 


45.1 
45.1 
45.1 
45.0 
45.0 
45.0 
44.9 
44.9 
44.9 
44.8 
44.8 
44.7 
44.7 
44.7 
44.6 
44.6 
44.6 
44.6 
44.5 
44.5 
44.5 
44.4 
44.4 
44.4 
44.3 
44.3 
44.3 
44.2 
44.2 
44.2 
44.1 
44.1 
44.0 
44.0 
44.0 
43.9 
43.9 
43.9 
43.8 
43.8 
43.8 
43.7 
43.7 
43.7 
43.7 
43.6 
43.6 
43.6 
43.5 
43.5 
43.5 
43.4 
43.4 
43.4 
43.3 
43.3 
43.3 
43.2 
43.2 
43.2 


9.8 
9.8 
9.8 
9.8 
9.8 
9.8 
9.9 
9.9 


Cosine.   ?D.  10" 

9.957276 

957217 

957158 

957099 

957040 

956981 

956921 

956862 

956803 

956744 

956684 
9.956625 

956566 

958598 

956447 

958387 

958327 

955268 

956208 

956148 

956089 

958029 

955969 

955909 

955849 

955789 

955729 

955S69 

95560!) 

955548 

955488 

955428 

955368 

955307 

955247 

955186 

955126 

955055 

955005 

954944 

954883 

9548-23 

954762 

954701 

954640 

954579 

954518 

954457 

954396 

954335 

954274 

954213 

954153 

954090 

954029 

953968 

953906 

953845 

953783 

953722 

953660 


9  9 

9  9 

9  9 

9  9 

9.9 

9.9 

9.9 

9.9 

9.9 

9.9 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.0 

10.1 

10 

10 


Tang. 


i 
1 

10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.1 
10.3 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.2 
10.3 


1.868673 
669002 
669332 
669661 
669991 
670320 
670849 
670977 
671308 
671634 
671983 

'.672291 
672619 
672947 
673274 
673602 
673929 
674257 
674584 
674910 
675237 

1.675564 
675890 
676216 
676543 
676859 
677194 
677520 
677846 
678171 
678496 

'•678821 
679146 
679471 
679795 
680120 
680444 
680768 
681092 
681416 
681740 

1.682083 
682387 
682710 
683033 
683356 
683679 
684001 
684324 
684646 
684968 

1.685290 
685612 
685934 
686255 
686577 
686898 
687219 
687540 
687861 
688182 

Cotang. 


D.  10' 


Ootanjj.  ;  N  .sine.  a.  cos 


10.331327! 
330998 ! 
330568 
330339 | 
330009 
339680 
329351 
329023 
328694 
328366 
328037 

10.327709 
327381 
327053 
326726 
326398 
326071 ! 
325743  j 
325416 | 
325090  j 
324763  i 

10.324436! 
324110 ' 
323784; 
323457 ! 
323131 | 
322806! 
322480 
333154 
321829; 
321504| 

10.3211791 
320854 | 
320529  i 
320205 | 
319880 | 
319556 j 
319232  I 
318908 
318584! 
318260 i 

10.317937! 
317613! 
317290 | 
316967] 
316644 
316321 
315999 
315676 
315354 
315032 | ! 

10.314710  j 
314388!! 
314036 | ! 
313745 ! 
313423 | i 
313102,! 
312781  || 
312460!j 
312139!| 
311818  | 


42262 
42288 
42315 
42341 
4236', 
42894 
42420 
42446 
42473 
42499 
42525 
42552 
42578 
42604 
42631 
4265; 
42683 
42709 
4273b 
42762 
42788 
42815 
42841 
42867 
42894 
42920 
42946 
42972 
42999 
43036 
43051 
43077 
43104 
43180 
43156 
43182 
43209 
43235 
43261 
43287 
43313 
43340 
43366 
43392 
43418 
43445 
43471 
43497 
43523 
43549 
43575 
43602 
43536 
43654 
43680 
43705 
4373o 
43755 
43785 
43811 
43837 


Tang.   II  N.  cos.  N.f 


90331 

yoeis 

90606 

90594 
90582 
90569 
90557 
90545 
90532 
90520 
90507 
90495 
90483 
90470 
90458 
90446 
90433 
90421 
90408 
90396 
90383 
90371 
90358 
90346 
90334 
90321 
90^0y 
90296 
90284 
90271 
90-259 
90246 
90233 
90221 
90208 
90196 
90183 
90171 
90158 
90146 
90183 
90120 
90108 
900i;5 
90082 
900/0 
90057 
90045 
90032 
90019 
90007 
89994 
89981 
89968 
89956 
89943 
89930 
89918- 
89905 
89892 
89879 


64  Degrees. 


■CAliLT?  IT. 
Sina. 


Log.  Sines  and  Tangents.  (-2CP)     Natural  Sinea 


D.  10"i  Cosine 


0 

1 
o 

3 
4 
5 
6 

8 
9 
10 
11 
12 
13 
14 
15 
16 

r, 

18 
19 
20 
21 

23 
24 
25 
26 

27 
u8 
29 
30 
31 
82 
33 
34 
35 
36 
3T 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
53 
51 
62 
53 
51 
55 
66 
57 
68 
59 
30 


). 04 1842 
042101 
642360 
642618 
642877 
643135 
643393 
643650 
643908 
644165 
644423 

). 644680 
644936 
645193 
645450 
645703 
645962 
646218 
646474 
646729 
648984 

). 647240 
647494 
647749 
648004 
648258 
648512 
648766 
649020 
649274 
649527 

1.649781 
650034 
650287 
650539 
650/92 
651044 
651297 
651549 
651800 
652052 

.652304 
652555 
652806 
653057 
653308 
653558 
653808 
654059 
654309 
654558 

.654808 
655058 
65530 1 
655556 
655805 
656054 
655392 
0 ':■-.-■  5  jl 
650793 
05/017 


L).  lo"' 


43.1 

43.1 

43.1 

43.0 

43.0 

43.0 

43.0 

42.9 

42.9 

42.9 

42.8 

42.8 

42.8 

42.7 

42.7 

42.7 

42.6 

42.6 

42.6 

42.5 

42.5 

42.5 

42.4 

42.4 

42.4 

42.4 

42.3 

42.3 

42.3 

42.2 

42.2 

42.2 

42.2 

42.1 

42.1 

42.1 

42.0 

42.0 

42.0 

41.9 

41.9 

41.9 

41.8 

41.8 

41.8 

41.8 

41.7 

41.7 

41.7 

41.6 

41.6 

41.6 

41.6 

41.5 

41.5 

41.5 

41.4 

41.4 

41.4 

41.3 

9.95 


953660 
953539 
953537 
953475 
953413 
953352 
953290 
953228 
953166 
953104 
953042 
952980 
952918 
952855 
952793 
952731 
952669 
952608 
952544 
952481 
952419 
.952356 
952294 
952231 
952168 
952108 
952043 
951980 
951917 
951854 
951791 
.951728 
951665 
951602 
951539 
951476 
951412 
951349 
951286 
951222 
951159 
•951096 
951032 
950968 
950905 
959841 
950778 
950714 
950650 
950586 
950522 
.950458 
95U394 
950380 
950366 
950202 
950138 
950074 
950010 
949945 
949881 
Sine. 


10.8 
10.3 
10.3 
10.3 
10.3 


10 

li) 

10 

10 

10 

10 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.4 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.5 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.6 

10.7 

10.7 

10.7 

10.7 

10.7 

10.7 

10.7 

10.7 

10.7 

10 


Tang. 

688182 
688502 
688823 
689143 
689463 
689783 
690103 
690423 
690742 
691062 
691381 
691700 
692019 
692338 
692656 
692975 
693293 
693612 
693930 
694248 
694566 
694883 
695201 
695518 
695836 
696153 
696470 
696787 
697103 
697420 
697736 
698053 
698369 
698685 
699001 
699316 
699632 
699947 
700263 
7005/8 
700893 
01208 
701523 
701837 
702152 
702466 
702180 
703095 
703409 
703/23 
704038 

).  704350 
704663 
704977 
705290 
705608 
705916 
706228 
703541 
706854 
707166 

*  ofcang. 


9.7 


P.  10' 

53.4 

53.4 

53.4 

53.3 

53.3 

53.3 

53.3 

53.3 

53.2 

53.2 

53.2 

53.1 

53.1 

53.1 

53.1 

63.1 

53.0 

53.0 

53.0 

53. 0 

52.9 

52.9 

52.9 

52.9 

52.9 

52.8 

52.8 

52.8 

62.8 

52.7 

62.7 

52.7 

62.7 

52.6 

52.6 

52.6 

52.6 

52.6 

52.5 

52.5 

62.5 

52.4 

52.4 

52.4 

52.4 

52.4 

52.3 

52.3 

52.3 

52.3 

52.2 

52.2 

52.2 

52.2 

52.2 

52.1 

52.1 

52.1 

52.1 

52.1 


Cotang.   N.  siue 

10.311818 

311498 

311177: 

310857  i 

8105371 

310217 ; 

309897 

309577 ! 

309258 ! 

308938  I 

308619  | 
10.308300! 

307981 ! 

307662  j 

307344 | 

307025  i 

306707 | 

306388 ! 

306070 | 

305752 I 

305434  j 
10.305117 | 

304799 

304482 | 

304164 

303847 

303530 

303213 

302897 

302580 

302264 
10-301947 

301631 

301315 

300999 

300684 

300368 

300053 

299737 

299422 

299107 
10-298792 

298477 

298163 

297848 

297534 

297220 

296905 

296591 

296277 

285964 
.295650 

295337 

295023 

294710 

294397 
294084 

293772 

293459 

293146 
292834 


10 


Tang. 


! 43837 
43863 

i  43889 
| 43916 
! 43942 
1 43968 
I  43994 
i 44020 
! 44046 
: 44072 
! 44098 
'44124 
; 44151 
! 44177 
; 44203 
44229 
44255 
44281 
44307 
44333 
44359 
44385 
44411 
44437 
44464 
44490 
44516 
44542 
44568 
44594 
44620 
44646 
44672 
44698 
44724 


N. cos. 

89879 
89867 
89854 
89841 
89828 
89816 
89803 
89790 
89777 
89764 
89752 
89739 
89726 
89713 
89709 
89687 
89674 
89662 
89649 
89636 
89623 
89610 
89597 
89584 
8957 1 
89558 
89545 
89532 
89519 
89508 
89493 
89480 
89467 
89454 
89441 


4475089428 
44776  89415 
4480289402 
44828  89389 


44854 
44880 
44906 
44932 
44958 
44984 
45010 
45036 
45062 
45088 
45114 
45140 
45166 
45192 
45218 
15243 
45269 
45295 
45321 
45347 
45373 
45599 


89376 
89363 
89350 
89337 
89324 
89311 
89298 
89285 
89272 
89259 
89245 
89232 
89219 
89206 
B9193 
89180 
89167 
89153 
8yl40 
89127 
89114 
89101 


:\.  cos.  N.sine 


60 
69 

58 
57 
56 
55 
54 
63 
52 
51 
50 
49 
4H 
47 
46 
45 
44 
48 
43 
41 
40 
89 
58 
37 
86 
85 
84 
83 
32 
81 
0 
29 
28 
27 
26 
26 
24 
23 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 
11 
10 
9 
8 
7 
6 
5 
4 
8 
2 
1 
0 


G3  Degrees. 


48 


Log.  Slues  and  Tangents.     (27°)     Natural  Sines. 


TABLE  II. 


Sine. 


D.  io 


0 

9.657047 

1 

657295 

2 

657542 

3 

657790 

4 

658037 

6 

658284 

6 

658531 

7 

658778 

8 

659025 

9 

659271 

10 

659517 

11 

9.659763 

12 

6601)09 

13 

660255 

14 

660501 

16 

660 743 

16 

660991 

1? 

661238 

is 

661481 

19 

661726 

80 

661970 

21 

9.662214 

22 

662459 

23 

662703 

24 

662946 

25 

663190 

26 

663433 

27 

663677 

28 

663920 

29 

664163 

30 

664406 

31 

9.664648 

32 

664891 

33 

665133 

34 

665375 

35 

665617 

36 

665859 

37 

666100 

CS 

666342 

39 

666583 

40 

666824 

41 

9.667055 

42 

667305 

43 

667546 

44 

667786 

45 

668027 

46 

668267 

47 

668506 

4S 

668746 

49 

668986 

54 

659225 

51 

9.669464 

52 

'  669703 

53 

669942 

54 

670181 

55 

670419 

56 

670658 

57 

670896 

58 

671134 

59 

671372 

60 

671609 

|  Cosine. 

41.3 
41.3 
41.2 
41.2 
41.2 
41.2 
41.1 
41.1 
41.1 
41.0 
41.0 
41.0 
40.9 
40.9 
40.9 
40.9 
40.8 
40.8 
40.8 
40.7 
40.7 
40.7 
40.7 
40.6 
40.6 
40.6 
40.5 
40.5 
40.5 
40.5 
40.4 
40.4 
40.4 
40.3 
40.3 
40.3 
40.2 
40.2 
40.2 
40.2 
40.1 
40.1 
40.1 
40.1 
40.0 
40  0 
40.0 
39.9 
39.9 
39.9 
39.9 
39.8 
39.8 
39.8 
39.7 
39.7 
39.7 
39.7 
39.6 
39.6 


9. 


949881 
949816 
949752 
949688 
949623 
949558 
949494 
949429 
949364 
949300 
949235 
949170 
949105 
949040 
948975 
948910 
948845 
948780 
948715 
948650 
948584 
948519 
948454 
948388 
948323 
948257 
948192 
948126 
948060 
947995 
947929 
947863 
947797 
947731 
947665 
947600 
947533 
947467 
947401 
947335 
947269 
947203 
947 \36 
947070 
947004 
946937 
946871 
946804 
946738 
946671 
946604 
946538 
946471 
946404 
946337 
946270 
946203 
946136 
946069 
946002 
945935 


10.7 
10.7 
10.7 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.8 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
10.9 
11.0 
11.0 
11.0 
11.0 
11.0 
11.0 
11.0 
11.0 
11.0 


11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11. 

11.2 

11.2 

11.2 

11.2 

11.2 


D.  n» 


.707166 

707478 

707790 

708102 

708414 

708726 

709037 

709349 

709660 

709971 

710282 
.710593 

710904 

711215 

711525 

711836 

712146 

712456 

712766 

713076 

713388 
.713696 

714005 

714314 

714624 

714933 

715242 

715551 

715860  if. 

716168  \l\  A. 

716477  &  * 
.716785 !&|  A. 

717093  I  "•* 

717401  I  b\-% 

717709!°  \'% 

718017  «•» 

718325  "-J 

718633  j" -| 

718940  « [l 

719248  "J  o 
719555 
.719862 
720169 
720476 


52.0 
52.0 
52.0 
52.0 
51.9 
51.9 
51.9 
51.9 
51.9 
51.8 
51.8 
51.8 
51.8 
51.8 
51.7 
51.7 
51.7 
51.7 
51.6 
51.6 
51.6 
51.6 
51.6 
51.5 
51.5 
51.5 
51.5 
51.4 


51.2 
51.2 
51.2 
51.1 


7204 /o--r 
720783 J}'} 
721089  I °}'J 
7213961°}-; 
721702  J}'* 
722009  °*" 
722315 ■?{•" 
722621  $H 
.722927 
723232 
723538 
723844 
724149 
724454 
724759 
725065 
725369 
725874 


51.0 
51.0 
50.9 
50.9 
50.9 
50.9 
50.9 
50.8 
50.8 
50.8 


Cotang. 


(Jotang.   N.  sine.  N.  cos. 

10.292834 

292522 

292210 

291898 

291586 

291274 

290963 

290651 

290340 

290029 

289718 
10.289407 

289096 

288785 

288475 

288164 

287854 

287544 

287234 

286924 

286614 
10.286304 

285995 

285686 

285376 

285067 

284758 

284449 

284140 

283832 

283523 
10.283215 

282907 

282599 

282291 

281983 

281675 

281367 

281060 

280752 

280445 
10.280138 

279831 

279524 
.279217 

278911 

278604 

278298 

277991 

277685 

277379 
10.277073 

276768 

276462 

276156 

275851 

275546 

275241 

274935 

274631 

274326 


45399  89101 

45425  89087 

I  4545189074 

45477  89061 

J  '45503  89048 

.i  45529  89035 

t  45554  89021 

.1145580  89008 

>  4560688995 

1:45632  88981 

1145658  88968 

45684  88955 

45710  88942 

45736  88928 

J  45762  88915 

1  45787  88902 

1145813  88888 

■ji  45839  88875 

145865  88862 

4589188848 

45917  88835 

45942  88822 

45968  88808 

45994  8S795 

46020  88782 

46046  88768 

46072  88755 

46097  88741 

46123  88728 

46149  88715 

46175  88701 

46201  88688 

46226  88674 

46252  88661 

46278  88647 

46304  88634 

46330  88620 

46365  88607' 

46381  88593 

46407  88580 

46433  88566 

46458  88553 

46484  88539 

46510  88526 

46536  88512 

46561  88499 

46587  88485 

46613  88472 

46639  88458 

46664  88445 

46696  88431 

46716  88417 

46742  88404 

46767  88390 

46793  88377 

46819  88363 

46844  88349 

46870  88336 

46896  88322 

46921  88308 

46947  8829f 


Tang. 


!  N.  cos. 


62  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.  (2S°)  Natural  Sines. 


49 


N.  sine. IN.  cos 


Siue. 


•  .671639 
671847 
672084 

67-2321 
672558 
672795 
673032 
673268 
673505 
673741 
673977 

'.674213 
674448 
674684 
674919 
675155 
675390 
675624 
675859 
676094 
676328 

'.676562 
676796 
677030 
677264 
677498 
677731 
677964 
678197 
678430 
678663 

.678895 
679128 
679360 
679592 
679824 
680356 
680288 
680519 
680750 
680982 

.681213 
681443 
681674 
681905 
682135 
682365 
682595 
682825 
683055 
683284 

.683514 
683743 
683972 
684201 
684430 
684658 
684887 
685115 
685348 
685571 


D.  10' 


39.6 

39.5 

39.5 

39 

39 

39 

39 

39 

39 

39 

39 

39 

39 

39.2 

39.2 

39.2 

39.1 

39.1 

39.1 

39.1 

39.0 

39.0 

39.0 

39.0 

38.9 

38.9 

38.9 

38.8 

38.8 

38.8 

38.8 

38.7 


38 
88 
38 
38 
38 

as 

38 

38.5 

38.5 

38.5 

38.4 

38.4 

38.4 

38.4 

38.3 

38.3 

38.3 

38.3 

38.2 

38.2 

38.2 

38.2 

38.1 

38.1 

38.1 

38.0 

38.0 

38.0 


Cosine. 


.945935 
945868 
945809 
945733 
945666 
945598 
945531 
945464 
945396 

•  945328 
945261 

.945193 
945125 
945058 
944990 
944922 
944854 
944786 
944718 
944650 
944582 

.944514 
944446 
944377 
944309 
944241 
944172 
944104 
944036 
943967 
943899 

.943830 
943761 
943693 
943624 
943555 
943486 
943417 
943348 
943279 
943210 

.943141 
943072 
943003 
942934 
942864 
942795 
942726 
942656 
942587 
942517 

.942448 
942378 
942308 
942239 
942169 
942099 
942029 
941959 
941889 
941819 
Sine. 


D.  10' 


Tang. 


725674 
725979 
726284 
726588 
726892 
727197 
727501 
727805 
728109 
728412 
728716 
72902C 
729323 
729626 
729929 
730233 
730535 
730838 
731141 
731444 
731746 

9.732048 
732351 
732653 
732955 
733257 
733558 
733860 
734162 
734463 
734764 

9.735066 
735367 
735668 
735969 
736269 
736570 
736871 
737171 
737471 
737771 
738071 
738371 
738671 
738971 
739271 
739570 
739870 
740169 
740468 
740767 

9.741066 
741365 
741664 
741962 
742261 
742559 
742858 
743156 
743454 
743752 
Cotang. 


D.  10" 


50.8 

50.8 

50  7 

50.7 

50.7 

50.7 

50.7 

50.6 

50.6 

50.6 

50.6 

50.6 

50-5 

50.5 

50.5 

50.5 

50.5 

50 

50 

50 

50 

50 

50 

50 

50 

50 

50 

50 

50-2 

50-2 

50-2 

50.2 

50.2 

50.1 

50.1 

50.1 

50.1 

50.1 

50.0 

50.0 

50.0 

50.0 

50.0 

49.9 

49.9 

49.9 

49.9 

49.9 

49.9 

49.8 

49.8 

49.8 

49.8 

49.8 

49.7 

49.7 

49.7 

49.7 

49.7 

49.7 


Co  tan; 


10.274326 
274021 
273716 
273412 
273108 
272803 
272499 
272195 
271891 
271588 
271284 

10.270980 
270677 
270374 
270071 
269767 
269465 
269162 
268859 
268556 
268254 

10.267952 
267649 
267347 
267045 
266743 
266442 
266140 
265838 
265537 
265236 

10.264934 
264633 
264332 
264031 
263731 
263430 
263129 
262829 
262529 
262229 

10.261929 
261629 
261329 
261029 
260729 
260430 
260130 
259831 
259532 
259233 

10.258934 
258635 
258336 
258038 
257739 
257441 
257142 
258844 
256546 
256248 


;!  46947  88295 
46973  88281 
4699988267 
4702488254 
4705088240 
47076  88226 
47101  88213 
47127  88199 
,.47153  88185 
"47178  88172 
4720488158 


47229 
47255 
47281 
47306 
47332 
47358 
47383 
47409 
47434 


4746088020 
47486  88006 
47511  87993 
47537  87979 
47562  87965 
47588  87951 
47614  87937 


88144 
88130 
88117 
88103 
88089 
88075 
88062 
88048 
88034 


47639 

47665 

47690 

i  47716 

: ! 47741 
47767 
47793 
47818 


87923 
87909 
87896 
87882 
87868 
87854 
87840 
87826 


47844J87812 
47869  87798 
47895  87784 
47920  87770 


|  47946 
ij  47971 
47997 
.48022 
;' 48048 
I  48073 
;!  48099 


87756 
87743 
87729 
87715 
87701 
87687 
87673 


148124  87659 
I  48150  87645 
(48175187631 


48201 


87617 


!  48226  87603 


: 48252 
| 48277 
! 48303 
: 48328 
I 48354 
148379 
! 48405 
! 48430 
1 48456 
48481 


Tans.   !  N.  cos.  N.sine 


87589 
87575 
87561 
87546 
87532 
87518 
87504 
87490 
87476 
87462 


61  Degrees. 


50 


Log.  Sines  and  Tangents.     (29°)    Natural  Sines. 


TABLE  II. 


Sine. 


0 

9.685571 

1 

685799 

2 

686027 

3 

686254 

4 

686482 

5 

686709 

6 

686936 

7 

687163 

8 

687389 

9 

687616 

10 

687843 

11 

). 688059 

12 

688295 

13 

688521 

14 

688747 

15 

688972 

16 

689198 

17 

689423 

18 

689648 

19 

689873 

20 

690098 

21 

3.690323 

22 

690j48 

23 

690772 

24 

690996 

25 

691220 

36 

691444 

27 

691668 

28 

691892 

29 

692115 

30 

692339 

31 

9.692562 

32 

692785 

33 

693008 

34 

693231 

35 

693453 

36 

693676 

37 

693898 

38 

694120 

39 

694342 

40 

694564 

41 

9.694786 

42 

695007 

43 

695229 

44 

695450 

45 

695671 

46 

695892 

47 

696113 

48 

696334 

49 

696554 

50 

696775 

51 

9.696995 

52 

697215 

53 

697435 

54 

697654 

55 

697874 

56 

698094 

57 

698313 

68 

698532 

69 

698751 

60 

698970 

D.  10"|    Cosine.    [D.  lu" 


38.0 
37.9 
37.9 
37.9 
37.9 
37.8 
37.8 
37.8 
37.8 
37.7 
37.7 
37.7 
37.7 
37.6 
37.6 
37.6 
37.6 
37.5 
37.5 
37.6 
37.5 
37.4 
37.4 
37.4 
37.4 
37.3 
37.3 
37.3 
37.3 
37.5 
37.2 
37.2 
37.1 
37.1 
37.1 
37.1 
37.0 
37.0 
37.0 
37.0 
36.9 
36.9 
36.9 
36.9 
36.8 
36.8 
36.8 
36.8 
36.7 
36.7 
36.7 
36.7 
36.6 
36.6 
36.6 
36.6 
36.5 
36.5 
36.5 
36.5 


Cosine. 


.941819 
941749 
941679 
941609 
941539 
941469 
941398 
941328 
941258 
941187 
941117 

.941046 
940975 
940905 
940834 
940763 
940693 
940622 
940551 
940480 
940409 

.940338 
940267 
940196 
940125 
940054 
939982 
939911 
939840 
939768 
939697 

.939625 
939554 
939482 
939410 

,939339 
939267 
939195 
939123 
939052 
938980 

.938908 
938836 
938763 
938691 
938619 
938547 
938475 
938402 
938330 
938258 

1.938185 
938113 
938040 
937967 
937895 
937822 
937749 
937676 
937604 
937531 


Sine. 


11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.7 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.8 
11.9 
11.9 
11.9 
11.9 
11.9 
11.9 
11.9 
11,9 
11.9 
11.9 
11.9 
11.9 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.0 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 
12.1 


Tang. 


.743752 
744050 
744348 
744645 
744943 
745240 
745538 
745835 
746132 
746429 
746726 

. 747023 
747319 
747616 
747913 
748209 
748505 
748801 
749097 
749393 
749689 

.749985 
750281 
750576 
750872 
751167 
751462 
751757 
752052 
752347 
752642 

.752937 
753231 
753526 
753820 
754115 
754409 
754703 
754997 
755291 
755585 

1.755878 
756172 
756465 
756759 
757052 
757345 
757638 
757931 
758224 
758517 

1.758810 
759102 
769395 
759687 
759979 
760272 
760564 
760856 
761148 
761439 


Co  tang. 


D.  10" 


49.6 
49.6 
49.6 
49.6 
49.6 
49.6 
49.5 
49.5 
49.5 
49.5 
49.5 
49.4 
49.4 
49.4 
49.4 
49.4 
49.3 
49.3 
49.3 
49.3 
49.3 
49.3 
49.2 
49.2 
49.2 
49.2 
49.2 
49.2 
49.1 
49.1 
49.1 
49.1 
49.1 
49.1 
49.0 
49.0 
49.0 
49.0 
49.0 
49.0 
48.9 
48.9 
48.9 
48.9 
48.9 
48.9 
48.8 
48.8 
48.8 
48.8 
48.8 
48.8 
48.7 
48.7 
48.7 
48.7 
48.7 
48.7 
48.6 
48.6 


Cotang. 

10.256248 
255950 
255652 
255355 
255057 
254760 
254462 
254165 
253868 
253571 
253274 

10.252977 
252681 
252384 
252087 
251791 
251496 
251199 
250903 
250607 
250311 

10.250015 
249719 
249424 
249128 
248833 
248538 
248243 
247948 
247653 
247358 

10.247063 
246769 
246474 
246180 
245885 
545591 
245297 
245003 
244709 
244415 

10.244122 
243828 
243535 
243241 
242948 
242655 
242362 
242069 
241776 
241483 

10.241190 
240898 
240605 
240313 
240021 
239728 
239436 
239144 
238852 
238561 


N.  sine.  X.  cos. 

87462 
87448 
87434 
87420 
87406 
87391 
87377 


87107 
.-  87093 
""'  87079 
87064 
87050 
87036 
87021 
87007 
86993 
49344J86978 
49S69|86964 
49394*86949 
4941986935 
49445  86921 
i  149470186906 
4949586892 
4952186878 
49546!86863 
4957186849 
49596  J86834 
49622J86820 
49647  86805 


49672 
49697 
49723 
49748 


86791 
86777 
86762 
86748 


49798J86719 
49824  86704 
49849  86690 


Tang. 


49874 
49899 
49924 
49950 


86675 
86661 
86646 
86632 


499-15:86617 
5000U|86603 


I  N.  cos. L\. Pint- 


60 

69 

58 
57 
56 
55 
54 
53 
52 
61 
50 
49 
4b 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
35 
34 
33 
32 
31 
30 
39 
28 
■21 
26 
25 
24 
23 
22 
21 
30 
19 
IS 
17 
16 
15 
14 
13 
12 
11 
10 

9 
8 
7 
6 
5 
4 
3 
2 
1 
0 


60  Degrees. 


Log.  Sines  aril  Tangents,    (30°)    Natural  Sines. 


51 


0 


s 

4 
5 
6 

7 
8 
9 

10 

11 

12 
13 

14 
16 
16 

1? 
18 

19 

•Jo 

■21 
J2 
23 
24 
25 
26 
27 
28 
25 
31 
31 
32 
33 
34 
36 
36 
37 
3^ 
30 
•10 
-41 
4-2 
43 
44 
45 
46 
4? 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
68 
50 

60 


iD.  10' 


9.698970  „, 
699189  ™ 
699497  I ^ 
699626  i  ™ 
699844 :  %* 
700082  *■ 
700280  \™  ■ 
700498 \%°- 
700716  1 5' 
700933  S 
701151  i^ 
.701388  ^ 
701585  il 
701802  ^ 
702019  ™' 
702236  H 
702452  ™- 
702669  %l 
702885  ^ 
703101  tl 
703317  * 

9.703533  Jg" 
703749  JJ' 
703984  g? 
704179  *?' 
704395  Jg' 
704810  |  r!?' 
704825  |*?" 
705040  I g 
705254 | J 
705469  I  J 

9.705683  *? 
705898  j?? 
706112  £ 
706326  JJ' 
705539  J5- 
706753  J?r 
f  0596  7  r^S 

707180  ;;?■ 

707393  *>' 
707603  £?■ 
"07819  Jg 

35 
35 
35 
36 
35, 
35 
35. 
35. 
35. 


9.7 


708032 
708245 
708458 
703670 
703882 
709094 
709305 
709518 
709730 
9.709941 
710153 
710J64 
710575 
710785 
71096  7 
711208  g 
711419  jj? 
711629  t? 
711839  do 


Cosine. 


35 
35. 

35 
35 


9.937531 
937458 
937385 
937312 
937238 
937165 
937092 
937019 
936946 
936872 
936799 

9.936725 
936652 
936578 
936505 
938431 
936357 
936284 
936210 
936136 
936062 
.935988 
935914 
935840 
935766 
935692 
935618 
935543 
935459 
935395 
935320 

9.935246 
935171 
935097 
935022 
934948 
934873 
934798 
934723 
934649 
934574 

9.934499 
934424 
934349 
934274 
934199 
934123 
934048 
933973 
933898 
933822 
.933747 
933671 
933596 
933520 
933445 
933369 
933293 
933217 
933141 
933056 


D.  10' 


12.1 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.2 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.3 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 
12.4 


12.4 
12.4 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.5 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 
12.6 


Tan  ^. 


1.761439 
761731 
762023 
762314 
762603 
762897 
763188 
763479 
763770 
784081 
764352 

'.764643 
764933 
765224 
765514 
765805 
766095 
766385 
768675 
766965 
767255 

. 767545 
767834 
768124 
768413 
768703 
768992 
769281 
769570 
769380 
770148 

.770437 
770726 
771015 
771303 
771592 
771880 
772168 
772457 
772745 
773033 

.773321 
773608 
773896 
774184 
774471 
774759 
775046 
775333 
775621 
775908 

.776195 
776482 
776769 
777055 
777342 
777628 
777915 
778201 
778487 
778774 


D.  10' 


Cotanjr. 


48.6 
48.6 
48.6 

48.6 
48.5 
48.5 
48.5 
48.5 
48.5 
48.5 
48.4 
48.4 
48.4 
48.4 
48.4 
48.4 
48.4 
48.3 
48.3 
48.3 
48.3 
48.3 
48.3 
48.2 
48.2 
48.2 
48,2 
48.2 
48.2 
48.1 
48.1 
48.1 
48.1 
48.1 
48.1 
48.1 
48.0 
48.0 
48.0 
48.0 
48.0 
48.0 
47.9 
47.9 
47.9 
47.9 
47.9 
47.9 
47.9 
47.8 
47.8 
47.8 
47.8 
47.8 
47.8 
47.8 
47.7 
47.7 
47.7 
47.7 


N.  sin 


10.235357  : 50277 
235037  jj  50302 
234776  |!  50327 
234488! '50352 


50377 
50103 
50428 
50453 
50478 
50503 
50528 
50553 
50578 


234195 
233905 
233615 
233325 
233035 
232745 

10.232455 
232166 
231876 
231587  II  50503 
231297  !|  50628 
231008  |50854 
230719  ij  50679 
230430;  50704 
230140  150729 
229852  j 50754 

10-229563j!50779 
229274 'I  50304 
228985  '50820 
228697  1150854 


228408 


50879 


228120  50904 
227832  50929 
227543  50954 
227255  j 50979 
226967!  51004 
10-226679  I  j  51029 
226392: 1 51054 
226104|;  51079 
225816  |  j  51104 
225529  511589  85941 
225241  !  1 51 154  35926 
224954  151179  85911 


X.  cos 

36603 
86588 
86573 
36559 
36544 
36530 
36515 
86501 
86486 
36471 
38457 
38442 
88427 
86413 
36398 
86384 
86369 
33354 
86340 
86325 
36310 
38295 
86281 
85266 
88251 
86237 
86222 
86207 
86192 
86178 
86163 
83148 
86133 
86119 
86104 
86089 
86074 
86059 
86045 
86030 
86015 
88000 
85985 
35970 
85955 


224667 
224379 | 
224092  i 
10-223805' 
223518  i 
223231 i 
222945  I 
222658  I 
222372 | 
222085 | 
221799 | 
221512 
221226  I 


Tang. 


51204 
51229 
51254 
51279 
51304 
51329 
51354 
51379 
51404 
51429 
51454 
51479 
fl504 


85898 
85881 
85868 
85851 
85836 
85821 
85808 
85792 
85777 
85762 
85747 
85732 
85717 
N.  cos.  N.sine. 


Degrees. 


21 


Log.  Sines  and  Tangents.     (31c)     Natural  Sines. 


0  ) 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 
11 
12 
13 
14 
16 
16 
17 
18 
19 

20  | 

21  9 
22 
23 
24 
25 
36 
27 
28 
29 
30  j 


Sine. 

.711839 
712050 
712260 
712409 
712679 
712889 
713098 
713308 
713517 
713726 
713935 

.714144 
714352 
714561 
714769 
714978 
715186 
715394 
715602 
715809 
716017 

.716224 
716432 
716639 
716846 
717053 
717259 
717466 
717673 
717879 
718085 

.718291 
718497 
718703 
718909 
719114 
719320 
719525 
719730 
719935 
720140 

.720345 
720549 
720754 
720958 
721162 
721366 
721510 
721774 
721978 
722181 
.722385 
722588 
722791 
722994 
723197 
723400 
723603 
723805 
724007 
724210 
Cosine. 


lD.  10" 


35.0 

35.0 

35.0 

34.9 

34.9 

34.9 

34.9 

34.9 

34.8 

34.8 

34.8 

34.8 

34.7 

34.7 

34.7 

34.7 

34.7 

34.6 

34.6 

34.6 

34.6 

34.5 

34.5 

34.5 

34.5 

34.5 

34.4 

34.4 

34.4 

34.4 

34.3 

34.3 

34.3 

34.3 

34 

34 

34 

34 

34 

34 

34 

34 

34 

34.0 

34.0 

34  0 

34.0 

34.0 

33.9 

33.9 

33.9 

33.9 

33.9 

33.8 

33-8 

33.8 

33.8 

33.7 

33.7 

33.7 


Uosme. 

.933036 
932999 
932914 
932838 
932762 
932685 
932609 
932533 
932457 
932380 
932304 

.932228 
932151 
932075 
931998 
931921 
931845 
931768 
931691 
931614 
931537 

.931460 
931383 
931306 
931229 
931152 
931075 
930998 
930921 
930843 
930766 

1.930688 
930611 
930533 
930456 
930378 
930300 
930223 
930145 
930067 
929989 

). 9299 11 
929833 
929755 
929677 
929599 
929521 
929442 
929364 
929286 
929207 

). 929129 
929050 
928972 
928893 
928815 
928736 
928657 
928578 
928499 
928420 
Sine.- 


D.  10' 


12.6 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.7 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.8 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
12.9 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.0 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 
13.1 


Tang. 


778774 
779030 
779346 
779332 
779918 
780203 
780489 
780775 
781060 
781346 
781631 
781916 
782201 
782480 
782771 
783056 
783341 
783626 
783910 
784195 
784479 
784764 
785048 
785332 
785616 
785900 
786184 
786468 
786752 
787036 
787319 
787603 
787886 
788170 
788453 
788736 
789019 
789302 
789585 
789868 
790151 
790433 
790716 
790999 
791281 
791563 
791846 
792128 
792410 
792692 
792974 
,793256 
793538 
793819 
794101 
794383 
794664 
794945 
795227 
795508 
795789 


D.  10' 


(Jotam 


47.7 
47.7 
47.6 
47.6 
47.6 
47.6 
47.6 
47.6 
47.6 
47.5 
47.5 
47.5 
47.5 
47.5 
47.5 
47.5 
47.5 
47.4 
47.4 
47.4 
47.4 
47.4 
47.4 
47.3 
47.3 
47.3 
47.3 
47.3 
47.3 
47.3 
47.2 
47.2 
47.2 
47.2 
47.2. 
47.2 
47.2 
47.1 
47.1 
47.1 
47.1 
47.1 
47.1 
47.1 
47.1 
47.0 
47.0 
47.0 
47.0 
47.0 
47.0 
47.0 
46.9 
46.9 
46.9 
46.9 
46.9 
46.9 
46.9 
46.8 


10 


Cotang. 
Degrees. 


10. 


10 


10 


10 


10 


221226 
220940 
220654 
220368 
220082 
219797 
219511 
219225 
218940 
218654 
218369 
218084 
217799 
217514 
217229 
216944 
216659 
216374 
216090 
215805 
215521 
215236 
214952 
214668 
214384 
214100 
213816 
213532 
213248 
212964 
212681 
212397 
212114 
211830 
211547 
211264 
210981 
210698 
210415 
210132 
209849 
209567 
209284 
209001 
208719 
208437 
208154 
207872 
207590 
207308 
207026 
.206744 
206462 
206181 
205899 
205617 
205336 
205055 
204773 
204492 
204211 


N.sine.jN.  co.s. 

51504185717 
51529(85702 
51554|85687 
51579185672 
51604)85657 
51628185642 
85627 


j 51653 
I  51678 
' 51703 

51728 
I  51753 

151778 
51803 


85612 
85597 
85582 
85567 
85551 
85536 
51828185521 


51852 
51877 
51902 
51927 


85506 
85491 
85476 
85461 


51952185446 
51977185431 
52002185416 
52026185401 
52051 185385 


52076 
52101 
52126 
52151 


85370 
85355 
85340 
85325 


52175185310 
52200185294 

52225J85279 
5225085264 
52275i85249 
52299J85234 


52324 
52349 
52374 
52399 
52423 
52448 
52473 


85218 
85203 
85188 
85173 
85157 
85142 
65127 


52498  85112 


52522 
52547 
52572 
52597 
52621 
52646 
52671 
52696 
52720 
52745 
52770 
152794 
i  52819 
i  52844 
1 52869 
1 52893 
| 52918 
1 52943 
|52967 
I 52992 


85096 
85081 
85066 
85051 
85035 
85020 
85005 
84989 
84974 
84959 
84943 
84928 
84913 
84897 
84882 
84866 
84851 
84836 
84820 
84805 


Tang.   ''  N.  cos.JN.aine,  ' 


TABLE  II. 


Log.  Sines  and  Tangents.    (32°)    Natural  Sines. 


53 


0 

l 
2 
S 
4 

6 
6 

7 
8 
9 
10 
11 
12 
13 
14 
16 
16 
17 
18 
19 
20 
•21 
2-2 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
84 
35 
86 
37 
38 

40 
41 
42 
43 
44 
45 
46 
47 
4S 
40 
50 
51 
52 
53 
54 
55 
50 
57 
68 
69 
60 


Sine. 


.724210 
724412 
724614 
724816 
725017 
725219 
725420 
725S22 
725823 
726024 
726225 
.726426 
726626 
726827 
727027 
727228 
727428 
727628 
727828 
728027 
728227 
.728427 
728626 
728825 
729024 
729223 
729422 
729621 
729820 
730018 
730216 
. 730415 
730613 
730811 
731009 
731208 
731404 
731602 
731799 
731996 
732193 
.  732390 
732587 
732784 
732980 
733177 
733373 
733569 
733765 
733961 
734157 
,734353 
734549 
734744 
734939 
735135 
735330 
735525 
735719 
735914 
736109 


Cosine. 


D.  10" 


33.7 
33.7 
33.6 
33.6 
33.6 
33.6 
33.5 
33.5 
33.5 
33.5 
33.5 
33.4 
33.4 
33.4 
33.4 
33.4 
33.3 
33.3 
33.3 
33.3 
33.3 
33.2 


33 

33 

33 

33 

33 

33 

33 

33.0 

33.0 

33.0 

33.0 

33.0 

32.9 

32.9 

32.9 

32.9 

32.9 

32.8 

32.8 

32.8 

32.8 

32.8 

32.7 

32.7 

32.7 

32.7 

32.7 

32.6 

32.6 

32.6 

32.6 

32.5 

32.5 

32.5 

32.5 

32.5 

32.4 

32.4 


Cosine.  |D.  10' 


9.928420 
928342 
928263 
928183 
928104 
928025 
927946 
927867 
927787 
927708 
927629 

9.927549 
927470 
927390 
927310 
927231 
927151 
927071 
926991 
92691 1 
926831 

9.926751 
926671 
926591 
926511 
926431 
926351 
926270 
926190 
926110 
926029 

9.925949 
925868 
925788 
925707 
925626 
925545 
925465 
925384 
925303 
925222 

9.925141 
925060 
924979 
924897 
924816 
924735 
924654 
924572 
924491 
924409 

9.924328 
924246 
924164 
924083 
924001 
923919 
923837 
923755 
923673 
923591 
"Sine.- " 


13.2 

13.2 

13.2 

13.2 

13.2 

13.2 

13.2 

13 

13 

13 

13 

13 

13 

13.3 

13.3 

13.3 

13.3 

13.3 

13.3 

13.3 

13.3 

13.3 

13.3 


13.3 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.4 
13.5 
13.5 
13.5 
13.5 
13.5 
13.5 
13.5 
13.5 
13.5 
13.5 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.6 
13.7 
13.7 


Tani?. 


9.795789 
796070 
796351 
796632 
796913 
797194 
797475 
797755 
798036 
798316 
798596 

9.798877 
799157 
799437 
799717 
799997 
800277 
800557 
800836 
801116 
801396 

9.801675 
801955 
802234 
802513 
802792 
803072 
803351 
803630 
803908 
804187 

9.804466 
804745 
805023 
805302 
805580 
805859 
806137 
806415 
806693 
806971 

9.807249 
807527 
807805 
808083 
808361 
808638 
808916 
809193 
809471 
809748 

9.810025 
810302 
810580 
810857 
811134 
811410 
811687 
811964 
■812241 
812517 

"Cotang. 


D.  10' 


46.8 
46.8 
46.8 
46.8 
46.8 
46.8 
46.8 
46.8 
46.7 
46.7 
46.7 
46.7 
46.7 
46.7 
46.7 
46.6 
46.6 
46.6 
46.6 
46.6 
46.6 
46.6 
46.6 
46.5 
46.5 
46.5 
46.5 
46.5 
46.5 
46.5 
46.5 
46.4 
46.4 
46.4 
46.4 
46.4 
46.4 
46.4 
46.3 
46.3 
46.3 
46.3 
46.3 
46.3 
46.3 
46.3 
46.2 
46.2 
46.2 
46.2 
46.2 
46.2 
46.2 
46.2 
46.2 
46.1 
46.1 
46.1 
46.1 
46.1 


Co  tail'. 


10.204211 
203930 
203649 
203368 
203087 
202806 
202525 
202245 
201964 
201684 
201404 

10.201123 
200843 
200563 
200283 
200003 
199723 
199443 
199164 
198884 
198604 

10.198325 
198045 
197766 
197487 
197208 
196928 
196649 
196370 
196092 
195813 

10.195534 
195255 
194977 
194698 
194420 
194141 
193863 
193585 
193307 
193029 

10.192751 
192473 
192195 
191917 
191639 
191362 
191084 
190807 
190529 
190252 

10.189975 


189420 
189143 
188866 
188590 
188313 
188036 
187759 
187483 


N.  s;iH\[_\.  COS. 

84805 
84789 
84774 
84759 
84743 
84728 
84712 
84697 
84681 
84666 
84650 
84635 
84619 


}  152992 
53017 
53041 
53036 
53091 
53115 
53140 
53164 
53189 
53214 
53238 
53263 
53288 


Tang. 


53312  84604 
!  53337  84588 
j  53361  84573 
53386J84557 
S53411J84542 
j  53435  84526 
5346084511 
!  53484  84495 
'  53509  84480 
!  53534  84464 
1 53558  84448 
53583  [84433 
53007:84417 
5363284402 
5365684386 
53681 184370 
;  53705!84355 
53730J84339 
53754184324 
I  53779184308 
j  53804184292 
|53828|84277 
|  53853J84261 
I  53877  [84245 
153902  '84230 
i  53926J84214 
163951 '84198 
J53975J84182 
!  54000  84167 
{ 54024!84151 
!  54049!84135 
I54073J84120 
154097184104 
1 54122^84088 
|  54146J84072 
5417184057 
54195:84041 
54220J84025 
5424484009 
54269  83994 
54293 183978 
54317183962 
54342 '83946 
54366J83930 
54S91  83915 
5441583899 
54440  83883 
54464  83867 


N.  cos.  N.sine. 


57  Degrees. 


54 


Log.  Sines  and  Tangents.     (33°)    Natural  Sines. 


TABLE  II. 


■me.      {P.  10 


0 
1 
2 
3 
4 
5 
6 
7 
8 
9 

id 

ii 

12 

13 
14 

16 
16 

17 

18 
1!) 

w 

21 
22 
23 
24 
25 
26 
21 
28 
2!) 
30 
31 
32 
33 
34 
35 
36 
37 
38 
89 
40 
41 
42 
4:1 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
64 
55 
54 
57 
58 
54 
60 


K 736109 

736303 
736498 
736692 
736886 
737080 
737274 
737467 
737661 
737855 
738018 

'.738241 
738434 
738027 
738820 
739013 
739208 
739398 
7395J0 
739783 
739975 

.740167 
740359 
740550 
740742 
740934 
741125 
741316 
741508 
741699 
741889 

.742080 
742271 
742462 
742652 
742342 
743033 
743223 
743413 
743602 
743792 

. 743982 
744171 
744361 
744550 
744739 
744928 
745117 
745305 
745494 
745I;83 

.745871 
746059 
746248 
746436 
746624 
746812 
746999 
747187 
747374 
747562 

Cosine. 


32.4 
32.4 
32.4 
32.3 
32.3 
32.3 
32.3 
32.3 
32.2 
32.2 
32.  2 
32.2 
32.2 
32.1 
32.1 
32.1 
32.1 
32.1 
32.0 
32.0 
32.0 
32.0 
32.0 
31.9 
31.9 
31.9 
31.9 
31.9 
31.8 
31.8 
31.8 
31.8 
31.8 
31.7 
31.7 
31.7 
31.7 
31.7 
31.6 
31.6 
31.6 
31.6 
31.6 
31.5 
31.5 
31.5 
31.5 
31.6 
31.4 
31.4 
31.4 
31.4 
31.4 
31.3 
31.3 
31.3 
31.3 
31.3 
31.2 
31.2 


.923591 
923509 
923427 
923345 
923263 
923181 
923098 
923016 
922933 
922851 
922768 

.922086 
922603 
922520 
922438 
922355 
922272 
922189 
922105 
922023 
921940 

.921857 
921774 
921691 
921607 
921524 
921441 
921357 
921274 
921190 
921107 

.921023 
920939 
920856 
920772 
920688 
920304 
920520 
920436 
920352 
921)268 

.920184 
920J99 
920015 
919931 
9lyS46 
919762 
919677 
919593 
919508 
919424 

.919339 
919254 
919169 
919085 
919000 
918915 
918830 
918745 
918659 
9)8574 


Sim-. 


L>.  lo' 

13.7 
13.7 
13.7 
13.7 
13.7 
13.7 
13.7 
13.7 
13.7 
13.7 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.8 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
13.9 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.0 
14.1 
14.1 
14.1 
14.1 
14.1 
14.1 
14.1 
14.1 
14.1 


14.1 
14.1 
14.1 
14.2 
14.2 
14.2 
14.2 


Tang. 

,812517 
812794 
813070 
813347 
813623 
813899 
814175 
814452 
814728 
815004 
815279 
815555 
815831 
816107 
816382 
816658 
816933 
817209 
817484 
817759 
818035 
818310 
818585 
818860 
819135 
819410 
819684 
819959 
820234 
820508 
820783 

9.821057 
821332 
821606 
821880 
822154 
822429 
822703 
822977 
823250 
823524 
823798 
824072 
824345 
824619 
824893 
825166 
825439 
825713 
825986 
826259 

9.826532 
826805 
827078 
827351 
827624 
827897 
828170 
828442 
828715 
828987 


Cotang. 


D.  10 

46.1 

46.1 

46  1 

46.0 

46.0 

46.0 

46.0 

46.0 

46.0 

46.0 

46.0 

45.9 

45.9 

45.9 

45.9 

45.9 

45. 9H 

45.9 

45.9 

45.9 

45.8 

45.8 

45.8 

45.8 

45.8 

45.8 

45.8 

45.8 

45.8 

45.7 

45.7 

45.7 

45.7 

45.7 

45.7 

45.7 

45.7 

45.7 

45.6 

45.6 

45.6 

45.6 

45.6 

45.6 

45.6 

45.6 

45.6 

45.5 

45.5 

45.5 

45.5 

45.6 

45.5 

45.5 

45.5 

45.5 

45.4 

45.4 

45.4 

45.4 


Cotang 


10 


10 


10 


10 


183893 | 
1836181 
183342 
183067 
182791 
1825161 
182241 ! 
181965  j 
.181696! 
181415! 
181140;! 
180865!; 
180590 !! 
180316  i' 
180041 j 
179766 h 
179492  j! 
1792171; 
,178943  ij 
178668 i ! 
178394 
178120 
177846 j 
177571 
177297 ' 
177023 
176750 
176476  ■ 
176202 
175928  i1 
175655H 
175381 | j 
175107! 
174834  ! 
174561 II 
174287 
174014 
173741 
173458 
173195 
172922 
172649 
172376 
172103 
171830 
171558 
171285 
171013 


54805:83645 
,54829  83629 
!!  54854  83613 
|j54878!83597 
'54902183581 
!  54927183565 
1  54951  ;83549 
54975  83533 
I54999J83517 
!  55024:83501 
55048:83485 
55072  !83469 
5509i!83453 
55121183437 
;55145!83421 
55169!83405 
55194i83389 
55218|83373 
55242  83o56 
j!  55266  83340 
I  55291  !83324 
■j55315!83308 
55339 183292 
55363  832  i  6 
55388  83260 
55412  83244 
55436;83228 
55460183212 
55484'83195 
55509;83179 
55533:83163 
55557!83147 
55581  !83131 
55605  ^831 15 
55630 '83008 
55654:83082 
i.5678!830.,0 
55702  83050 
55726:83034 
55750'83017 
55775183001 
55799I82L85 
55823  82669 
82953 
82936 
82920 


55847 
55871 
55895 


Tarn 


55919182904 
N.  cos.JN.sine. 


56  Degrees. 


Lop;.  Sines  and  Tangents.     (34°)     Natural  Sines. 


55 


0 

2 
3 
4 

5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
1? 
is 
19 
20 
21 
22 
23 
24 
25 
36 
2? 
28 
29 
80 
81 
32 
33 
84 
35 
86 
37 
88 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
66 
57 
58 
59 
60 


.747502 
747749 
747936 
748123 
748310 
748497 
748683 
748870 
749056 
749243 
749426 

.749815 
749801 
749987 
750172 
750358 
750543 
750729 
750914 
751099 
751284 

.751469 
751654 
751839 
752023 
752208 
752392 
752576 
752760 
752944 
753128 

.753312 
753495 
753679 
753862 
754046 
754229 
754412 
754595 
754778 
754960 
755143 
755326 
755508 
755690 
755872 
756054 
756236 
756418 
756600 
756782 

.756963 
757144 
757326 
757507 
757688 
757869 
758050 
758230 
758411 

J758591 

Cosine. 


D.  10" 


9.7 


31 

31 

31 

31 

31 

31 

31 

31.1 

31.0 

31.0 

31.0 

31.0 

31.0 

30.9 

30.9 

30.9 

30.9 

30.9 

30.8 

30.8 

30.8 

30.8 

30.8 

30.8 

30.7 

30.7 

30.7 

30.7 

30.7 

30.6 

30.6 

30.6 

30.6 

30.6 

30.5 

30.5 

30.5 

30  5 

30.5 

30.4. 

30.4 

30.4 

30.4 

30.4 


30.2 


Cosine. 

1.918574 
918489 
918404 
918318 
918233 
918147 
918032 
917976 
917891 
917805 
917719 

1.917634 
917548 
917462 
917376 
917290 
917204 
917118 
917032 
916946 
916859 

.916773 
916687 
916600 
916514 
916427 
916341 
916254 
916167 
916081 
915994 

.915907 
915820 
915733 
915646 
915559 
915472 
915385 
915297 
915210 
915123 

.915035 
914948 
914860 
914773 
914685 
914598 
914510 
914422 
914334 
914246 

.914158 
914070 
913982 
913894 
913806 
913718 
913630 
913541 
913453 
913365 

~SimT~ 


D.  10" 


14.2 
14.2 
14.2 
14.2 
14.2 
14.2 
14.2 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.3 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.4 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.5 
14.6 
14.6 
14.6 
14.6 
14.6 
14.6 
14.6 
14.6 
14.6 
14.6 
14.7 
14.7 
14.7 
14.7 
14.7 
14.7 
14.7 
14.7 
14.7 
14.7 


9. 


Tang. 

.828987 
829260 
829532 
829805 
830077 
830349 
830621 
830893 
831165 
831437 
831709 
.831981 
832253 
832525 
832796 
833068 
833339 
833611 
833882 
834154 
834425 
834696 
834967 
835238 
835509 
835780 
836051 
836322 
836593 
836864 
837134 

►.837405 
837675 
837946 
838216 
838487 
838757 
839027 
839297 
839568 
839838 

1.840108 
840378 
840647 
840917 
841187 
841457 
841726 
841996 
842266 
842535 

1.842805 
843074 
843343 
843612 
843882 
844151 
844420 
844689 
844958 
845227 

Cotang. 


D.  10" 


45.4 
45.4 
45.4 
45.4 


Cotang.  jN. sine  X.  cos. 


45.4 

45.3 

45.3 

45.3 

45.3 

45.3 

45.3 

45.3 

45.3 

45 

45 

45 

45 

45 

45 

45 

45 

45 

45 

45 

45 

45.1 

45.1 

45.1 

45.1 

45.1 

45.1 

45.1 

45.1 

45.1 

45.1 

45.0 

45.0 

45.0 

45.0 

45.0 

45.0 

45.0 

45.0 

45.0 

44.9 
44.9 
44.9 
44.9 
44.9 
44.9 
44.9 
44.9 
44.9 
44.8 
44.8 
44.8 
44.8 
44.8 


10 


10 


10 


10 


10 


171013 
170740 
170468 
170195 
169923 
169651 
169379 
169107 
168835 
168563 
168291 
168019 
167747 
167475 
167204 
166932 
166661 
166389 
166118 
165846 
165575 
165304 
165033 
164762 
164491 
164220 
163949 
163678 
163407 
163136 
162866 
162595 
162325 
162054 
161784 
161513 
161243 
160973 
160703 
160432 
160162 
159892 
159622 
159353 
159083 
158813 
158543 
158274 
158004 
157734 
157465 
157195 
156926 
156657 
156388 
156118 
155849 
155580 
155311 
155042 
154773 

Ting- 


: 55919 
| 55943 

55968 
! 55992 
I  56016 
! 56040 
J  56064 

56088 
156112 
| 56136 
| 56160 

56184 


32904 

8288' 
82871 
82855 
82839 
82822 
82806 
82790 
82773 
82757 
82741 
82724 


56208  82708 


56232 
56256 
56280 
56305 
56329 
56353 
56377 
56401 
56425 
56449 
56473 
56497 
56521 
56545 
56569 
56593 
56617 
56641 
56665 
56689 
56713 
56736 
56760 
56784 
56808 
56832 
56856 
56880 
56904 
56928 
56952 
56976 
57000 
57024 
57047 
57071 
57095 
57119 
57143 
57167 
57191 
57215 
57238 
57262 
57286 
57310 
57334 
57358 


82692 
82675 
82659 
82643 
82626 
82610 
82593 
82577 
82561 
82544 
82528 
82511 
82495 
82478 
82462 
82446 
82429 
82413 
82396 
82380 
82363 
82347 
82330 
82314 
82297 
82281 
82264 
82248 
82231 
82214 
82198 
82181 
82165 
82148 
82132 
82115 
82098 
82082 
82085 
82048 
82032 
82015 
81999 
81982 
81965 
81949 
81932 
81915 


X.  cos.  X  sine, 


55  Decrees. 


5G 


Log.  Sines  and  Tangents.    (35°)    Natural  Sines. 


TABLE  II. 


jsttne. 

9.758591 
758772 
758952 
759132 
759312 
759492 
759672 
759852 
760031 
760211 
760390 

9.760569 
760748 
760927 
761106 
761285 
761464 
761642 
761821 
761999 
762177 

9.762356 
762534 
762712 
762889 
763067 
763245 
763422 
763600 
763777 
763954 

9.764131 
764308 
764485 
764662 
764838 
765015 
765191 
766367 
765544 
765720 

9.765896 
766072 
766247 
766423 
766598 
766774 
766949 
767124 
767300 
767475 

9.767649 
767824 
767999 
768173 
768348 
768522 
768697 
768871 
769045 
769219 
Cosine. 


D.  10' 


30.1 

30.0 

30.0 

30  0 

30.  o 

30.0 

29.9 

29.9 

29.9 

29.9 

29.9 

29.8 

29.8 

29.8 

29.8 

29.8 

29.8 

29.7 

29.*? 

29.7 

29.7 

29.7 

29.6 

29.6 

29.6 

29.6 

29.6 

29.6 

29.5 

29.5 

29.5 

29.5 

29.5 

29.4 

29.4 

29.4 

29.4 

29.4 

29.4 

29.3 

29.3 

29.3 

29.3 

29.3 

29.3 

29.2 

29.2 

29.2 

29 

20 

■29 

■29 

29 

■29 

29 

29.0 

29.0 

29.0 

29.0 

29.0 


Cosine. 

.913365 
913276 
913187 
913099 
913010 
912922 
912833 
912744 
912655 
912566 
912477 

.912388 
912299 
912210 
912121 
912031 
911942 
911853 
911763 
911674 
911584 

.911495 
911405 
911315 
911226 
911136 
911046 
910956 
910866 
910776 
910686 

.91U596 
910506 
910415 
910325 
910235 
910144 
910054 
909963 
909873 
909782 

.909691 
909601 
909510 
909419 
909328 
909237 
909146 
909055 
908964 
908873 

1.908781 
908690 
9035-9 
908507 
908416 
908324 
908233 
908141 
903049 
907958 
Sine. 


U.  10" 


4 

4 

4 
4 

4 
4 

4.8 

4.8 

4.8 

4.8 

4.8 

4 

4 


4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
4.9 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.0 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.1 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.2 
5.3 
5.3 
5 . 3 
5.3 
6 . 3 


Tanir. 


.845227 
845496 
845764 
846033 
846302 
846570 
846839 
847107 
847376 
847644 
847913 

'.848181 
848449 
848717 
848986 
849254 
849522 
849790 
850058 
850325 
850593 

.850861 
851129 
851396 
851664 
851931 
852199 
852466 
852733 
853001 
853268 

.853535 
853802 
854069 
854336 
854603 
854870 
855137 
855404 
855671 
855938 

'.856204 
856471 
856737 
857004 
857270 
857537 
857803 
858069 
858336 
858602 

1.858868 
859134 
859400 
859666 
859932 
860198 
860464 
860730 
860995 
861261 

Cotang. 


44.8 
44.8 
44.8 
44.8 
44.8 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.7 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.6 
44.5 
44.6 
44.5 
44.5 
44.5 
44.5 
44.5 
44.5 
44.5 
44.5 
44.5 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.4 
44.3 
44.3 
44.3 
44.3 
44.3 
44.3 
44.3 
44.3 
44.3 
44.3 


Cotang.  I  N.  ?ine.  N.  cos. 


10.154773 
154504 
154236 
153987 
153698 
153430 
153161 
152893 
152624 
152356 
152087 

10.151819 
151551 
151283 
151014 
150746 
150478 
150210 
149942 
149675 
149407 

10-149139 
148871 
148604 
148336 
148089 
147801 
147534 
147267 
146999 
146732 

10-146465 
146198 
145931 
145664 
145397 
145130 
144863 
144596 
144329 
144062 

10-143796 
143529 
143263 
142996 
142730 
142463 
142197 
141931 
141664 
141398 

10-141132 
140866 
140600 
140334 
140068 
139802 
139536 
139270 
139005 
138739 
Tang. 


I  57358 
j;  57381 
I;  57405 

I  57429 
[  57453 

I I  57477 
i  57501 

i  57524 
57548 
57572 

< 67596 


81915 
81899 
81882 
81865 
81848 
81832 
81815 
81798 
81782 
81765 
81743 


|  57643  81714 
|57667  81698 
57691 181681 
57715J81664 
67738181647 
57762  81631 
,57786  81614 
!57810!81597 
157833  81580 
57857181563 
5788181546 
5790481530 
5792881513 
57952l81496 
57976J81479 
|  57999,81462 
58023;81445 
I  58047181428 
158070  81412 


58094 
58118 
58141 
58165 
58189 
58212 
58236 
58260 


81395 
81378 
81361 
81344 
81327 
81310 
81293 
HI  276 


58283  81259 


58307 
58330 
58354 
58378 
58401 
58425 
58449 
58472 
58496 
58519 
58543 
I  58567 
1 58590 
I  58614 


81242 
81225 
81208 
81191 
81174 
81157 
81140 
81123 
81106 
81039 
81072 
81055 
81038 
81021 


58637 \s 1004 


! 58661 

! 58684 
j  58708 
158731 
! 58755 
i 58 779 


80987 
80370 
80953 
30036 

80019 
809U2 


N.  C06.  -'•..? 


54  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (36°)    Natural  Sines. 


57 


0 

9.769219 

1 

769393 

2 

769566 

3 

769740 

4 

769913 

5 

770087 

6 

770260 

7 

770433 

6 

770606 

9 

770779 

10 

770952 

11 

9.771125 

13 

771298 

13 

771470 

11 

771643 

16 

771815 

16 

771987 

17 

772159 

18 

772331 

19 

772503 

30 

772675 

•21 

9.772847 

22 

773018 

■23 

773190 

24 

773361 

25 

773533 

26 

773704 

27 

773875 

28 

774046 

29 

774217 

30 

774388 

31 

9.774558 

32 

774729 

33 

774899 

31 

775070 

35 

775240 

36 

775410 

:)7 

775580 

38 

775750 

39 

775920 

40 

776090 

41 

9.776259 

42 

776429 

43 

776598 

44 

776768 

45 

776937 

46 

777108 

47 

777275 

48 

777444 

49 

777613 

50 

777781 

51 

9.777950 

52 

778119 

53 

778287 

54 

778455 

55 

778824 

56 

778792 

57 

778960 

58 

779128 

69 

779295 

60 

779483 

Cosine. 


D.  10"     Cosine. 


29.0 

28.9 

28.9 

28.9 

28.9 

28.9 

28.8 

28.8 

28.8 

28.8 

28.8 

28.8 

28.7 

28.7 

28.7 

28.7 

28.7 

28.7 

28.6 

28.6 

28.6 

28.6 

28.6 

28.6 

28.5 

28.5 

28.5 

28.5 

28.5 

28.5 

28.4 

28.4 

28.4 

28.4 

28.4 

28.4 

28.3 

28.3 

28.3 

28.3 

28.3 

28 

28 

28 

28 

28 

28 

28 

28 

28 

28 

28 

28.1 

28.0 

28.0 

28.0 

28.0 

28.0 

28.0 

27.9 


.907958 
907866 
907774 
907682 
907590 
907498 
907406 
907314 
907222 
907129 
907037 

.908945 
906852 
906760 
906667 
906575 
906482 
906389 
906296 
906204 
906111 

.906018 
905925 
905832 
905739 
905645 
905552 
905459 
905366 
905272 
905179 

.905085 
904992 
904898 
904804 
904711 
904617 
904523 
904429 
904335 
904241 

.904147 
904053 
903959 
903864 
903770 
903676 
903581 
903487 
903392 
903298 

.903202 
903108 
903014 
902919 
902824 
902729 
902634 
902539 
902444 
902349 


Sine. 


D.  10"      Tang.      D.  10 


15.3 
15.3 
15.3 
15.3 
15.3 
15.3 
15.3 
15.4 
15.4 
15.4 


15 

15 

15 

15 

15 

15 

15.4 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.5 

15.6 

15.6 

15.6 

15.6 

15.6 

15.6 

15.6 

15.6 

15.6 

15.6 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.7 

15.8 

15.8 

15.8 

15.8 

15.8 

15.8 

15.8 

15.8 

15.8 

15.9 

15.9 


9. 


.861261 
861527 
861792 
862058 
862323 
862589 
862854 
863119 
863385 
863650 
863915 

.864180 
864445 
864710 
864975 
865240 
865505 
865770 
866035 
866300 
866564 

.866829 
867094 
867358 
867623 
867887 
868152 
868416. 
868680 
868945 
869209 

.869473 
869737 
870001 
870265 
870529 
870793 
871057 
871321 
871585 
871849 
872112 
872376 
872640 
872903 
873167 
873430 
873694 
873957 
874220 
874484 
874747 
875010 
875273 
875536 
875800 
876063 
876326 
876589 
876851 
877114 


Co tan c 


44.3 

44.3 

44  2 

44.2 

44.2 

44.2 

44.2 

44.2 

44.2 

44.2 

44.2 

44 

44 

44 

44 

44 

44 

44 

44 

44 

44 

44 

44.1 

44.1 

44.1 

44.1 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

44.0 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.9 

43.8 

43.8 

43.8 

43.8 

43.8 

43.8 

43.8 


Cotang. 


10.138739 
138473 
138208 
137942 
137677 
137411 
137146 
136881 
136615 
136350 
136085 

10.135820 
135555 
135290 
135025 
134760 
134495 
134230 
133965 
133700 
133436 

10.133171 
132906 
132642 
132377 
132113 
131848 
131584 
131320 
131055 
130791 

10.130527 
130263 
129999 
129735 
129471 
129207 
128943 
128679 
128415 
128151 

10.127888 
127624 
127360 
127097 
126833 
126570 
126306 
126043 
125780 
125516 
125253 
124990 
124727 
124464 
124200 
123937 
123674 
123411 
123149 
122886 
Tang. 


N.  sine.  N.  cos 


58779  80902 
58802  80885 
58826J80867 
5884980850 
5887380833 
58896,80816 
58920  80799 
58943  80782 
58967  80765 
58990  80748 
59014180730 
59037180713 
5906180696 
5908480679 
59108|80662 
59131)80644 
5915480627 
5917880610 
5920180593 
59225  80576 
59248  80558 
5927280541 
59295!80524 
59318!80507 
59342  80489 
59366 180472 
59389)80455 
59412J80438 
59436:80422 
59459|80403 
59482:80386 
59606180368 
69529|80361 
59552J80334 
59576  80316 
59599!80299 
59622  J80282 
59646  J80264 
59669  [80247 
59693  80230 
5971680212 
59739  80195 
59763J80178 
59786|80160 
59809|80143 
59832  8012 


10 


59856 
59879 
59902 
59926 
59949 


80108 
80091 
80073 

80056 

80038 


69972  J80021 
|  59995  80003 
60019179986 

!  60042|79968 


I  60065 
: 60089 
j 60112 
60135 
160158 
160182 


!  N.  cos*  N.sine 


79951 
?9934 
,"9916 
?9899 
J  9881 
79864 


53  Degrees. 


58 


Log.  Sines  and  Tangents.    (37°)    Natural  Sines. 


TABLE  II. 


0 

1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
Ifl 
20 
21 
22 
23 
24 
25 
26 
2  7 
28 
20 
39 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
5L 
52 
53 
54 
55 
56 
57 
58 
50 
00 


Sine. 

1.779463 
779631 
779798 
779966 
780133 
780300 
780467 
780534 
780801 
780968 
781134 

.781301 
781468 
781634 
781800 
781966 
782132 
782298 
782464 
782630 
782796 

.782961 
783127 
783282 
783458 
783623 
783788 
783053 
784118 
784282 
784447 

.784612 
784776 
784941 
785105 
785269 
785433 
785597 
785761 
785925 
786089 

. 786252 
786416 
786579 
786742 
786905 
787069 
787232 
787395 
787557 
787720 

.787883 
788045 
788208 
788370 
788532 
788694 
788856 
789018 
789180 
789342 
Cosine. 


D.  10' 


27.9 
27.9 
27.9 
27.9 
27.9 
27.8 
27.8 
27.8 
27.8 
27.8 
27.8 
27.7 
27.7 
27.7 
27.7 
27.7 
27.7 
27.6 
27.6 
27.6 
27.6 
27.6 
27.6 
27.5 
27.5 
27.5 
27.5 
27.5 


27 

27 

27 

27 

27 

27 

27 

27.3 

27.3 

27.3 

27.3 

27.3 

27.3 

27.2 

27.2 

27^2 

27.2 

27.2 

27.2 


27 

27 

27 

27 

27 

27 

27 

27.0 

27.0 

27.0 

27.0 

27.0 

27.0 


Cosine.  |D.  10' 

1.902349 

902253 

902158 

902063 

901967 

901872 

901776 

901681 

901585 

901490 

901394 
•.901298 

901202 

901106 

901010 

900914 

900818 

900722 

900626 

900529 

900433 
'.900337 

900242 

900144 

900047 

899951 

899854 

899757 

899660 

899584 

899467 
'.899370 

899273 

899176 

899078 

898981 

898884 

898787 

898689 

898592 

898494 
1.898397 

898299 

898202 

898104 

898006 

897908 

897810 

897712 

897614 

897516 
1.897418 

897320 

897222 

897123 

897025 

896926 

896828 

896729 

896631 

895532 
Sine. 


Tans. 


9.877114 
877377 
877640 
877903 
878165 
878428 
878691 
878953 
879216 
879478 
879741 

9.880003 
880265 
880528 
880790 
881052 
881314 
881576 
881839 
882101 
882363 

9.882625 
882887 
883148 
883410 
883672 
883934 
884196 
884457 
884719 
884980 

9.885242 
885503 
885765 
886026 
885288 
886549 
886810 
887072 
887333 
887594 

9.887855 
888116 
888377 
888639 
888900 
889160 
889421 
889682 
889943 
890204 
1.890465 
890725 
890986 
891247 
891507 
891768 
892028 
892289 
892549 
892810 
Cotang. 


D.  10" 


Cotang. 


10.122886 
122623 
122360 
122097 
121835 
121572 
121309 
121047 
120784 
120522 
120259 

10.119997 
119735 
119472 
119210 
118948 
118686 
118424 
118161 
117899 
117637 

10.117375 
117113 
116852 
116590 
116328 
116056 
115804 
115543 
115281 
115020 

10.114758 
114497 
114235 
113974 
113712 
113451 
113190 
112928 
112667 
112405 

10.112145 
111884 
111623 
111361 
111100 
110840 
110579 
110318 
110057 
109796 

10.109535 
109275 
109014 
108753 
108493 
108232 
10  i  9  72 
107711 
107451 
107190 


jN.sine 

60182 
60205 
60228 
60251 
60274 
60298 
60321 
60344 
60367 
60390 
60414 
60437 
60460 
60483 
60506 
60529 
60553 
60576 
60599 
60622 
60645 
60668 
60691 
60714 
60738 
60761 
60784 
6080 
60830 
60853 
60876 
60899 
60922 
60945 
60968 
60991 
61015 
61038 
61061 
61084 
61107 
61130 
61153 
61176 
61199 
61222 
61245 
i 61268 
i 61291 
161314 
61337 
161360 
161383 
! 61406 
! 61429 
161451 
j 61474 
161497 
| 61520 
61543 
; 6156b 


N.  cos 


79864 
79846 
79829 
79811 
79793 
79776 
79758 
79741 
79723 
79706 
79688 
79671 
79658 
79635 
79618 
79600 
79583 
79565 
79547 
79530 
79512 
79494 
79477 
79459 
79441 
79424 
79406 
79388 
79371 
79353 
79335 
79318 
79300 
79282 
79264 
79247 
79229 
79211 
79193 
.9176 
79158 
79140 
79122 
79105 
79087 
79069 
79051 
,9033 
79016 
;8998 
78980 
78962 
78944 
78926 
78908 
78891 
78873 
78855 
78837 
78819 
78801 


Tang,   h  N.  eo8.jy.8toe 


52  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (08°)    Natural  Sines. 


59 


0 

l 
3 

3 
4 

5 
6 

7 

8 

9 
10 
11 
12 
13 
14 
15 
16 
I? 
lb 
19 
20 
•21 
22 
33 
24 
25 
26 
27 
28 
29 
30 
31  \9 
32 
33 
34 
35 
30 
37 
38 
39 
40 
41 
42 
43 
44 
45 
45 
47 
48 
49 
50 

51 
5-2 
53 
54 
55 
56 
5  7 
58 
59 
60 


789342 
789504 
789665 
789827 
789988 
790149 
790310 
790471 
790832 
799793 
790954 
791115 
791275 
791436 
791596 
791757 
791917 
792077 
792237 
792397 
792557 
,792716 
792876 
793035 
793195 
793354 
793514 
793673 
793832 
793991 
794150 
.794308 
794467 
794626 
794784 
794942' 
795101 
795259 
795417 
795575 
795733 
.795891 
796049 
790206 
790364 
796521 
790679 
796836 
796993 
797150 
797307 
.797464 
797021 
797777 
797934 
798091 
798247 
798403 
79S560 
798716 
798872 


D,  10' 


26.9 
26.9 
26.9 
26.9 
26.9 
26.9 
26.8 
26.8 
26.8 
26.8 
28.8 
28.8 
26.7 
26.7 
26.7 
26.7 
26.7 
26.7 
26.6 
26.6 
28.6 
26.6 
26.6 
26.6 
28.5 
26.5 
28.5 
26.5 
26.5 
26.5 
26.4 
26.4 
26.4 
26.4 
26-4 
26-4 


26 

20 

20 

20 

20 

20 

20 

25 

20 

20 

20 

20 

26 

26-1 

26-1 


26 

20 
26 

20 

20 

26 

26.0 

20.0 

20.0 


Cosine. 


Cosine. 

.898532 
898433 
896335 
896238 
896137 
896038 
895939 
895840 
895741 
895641 
895542 

1.895443 
895343 
895244 
895145 
895045 
894945 
894846 
894746 
894646 
894546 

'.894446 
894346 
894246 
894146 
894046 
893946 
893846 
893745 
893645 
893544 

1.893444 
893343 
893243 
893142 
893041 
892940 
892839 
892739 
892638 
892536 

1.892435 
892334 
892233 
892132 
892030 
891929 
891827 
891726 
891624 
891523 

>.  891421 
891319 
891217 
891115 
8910U 
890911 
890809 
890707 
890505 
890503 


I).  ior; 


10 

10 

10 

10 

10 

16 

16.5 

16.5 

16.5 

16.5 

16.5 

16.6 

16.6 

16.6 

16.6 

16.6 

16.6 

16.6 

16.6 

16.6 

16.6 

16.7 

16.7 

16.7 

16.7 

16.7 

16.7 

16.7 

16.7 

16.7 

16.7 

16.8 

16.8 

16.8 

16 

10 

10 

16 

16 

16.8 

16.8 

16.9 

16.9 

16.9 

16.9 

16.9 

16.9 

16.9 

16.9 

16.9 

17.0 

17.0 

17.0 

17..  0 

17.0 

17.0 

17.0 

17.0 

17.0 

17.0 


Tani 


.£92310 
893070 
893331 
893591 
893851 
894111 
894371 
894632 
894892 
895152 
895412 

.895672 
895932 
896192 
896452 
896712 
896971 
897231 
897491 
897751 
898010 

.898270 
898530 
898789 
899049 
899308 
899563 
899827 
900086 
900340 
900605 

.900864 
901124 
901383 
901042 
901901 
902100 
902419 
902679 
902938 
903197 

1.903455 
903714 
903973 
904232 
904491 
904750 
905008 
905267 
905526 
905784 

(.900043 
900302 
900500 
906819 
907077 
907330 
907594 
907852 
908111 
908369 

Cotang. 


D.  10'- 

43.4 
43.4 
43.4 
43.4 
43.4 


43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43.3 

43.3 

43.3 

43.3 

43.3 

43.3 

43.3 

43.3 

43.2 

43.2 

43.2 

43.2 

43.2 

43.2 

43.2 

43.2 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.1 

43.0 


Cotang. 

10.107190 
106930 
106609 
100409 
106149 
105889 
105629 
105368 
105108 
104848 
104588 

10.104328 
104088 
103808 
103548 
103288 
103029 
102769 
102509 
102249 
101990 

10.101730 
101470 
101211 
100951 
100692 
100432 
100173 
099914 
099654 
099395 

10.099136 
098876 
098617 
098358 
098039 
097840 
097581 
097321 
097062 
096803 

10.090545 
090280 
096027 
095708 
095509 
095250 
094992 
094733 
094474 
094216 

10.093957 
093698 
093440 
093181 
092923 
092664 
092406 
092148 
091889 
091031 


N.  sine.iN.  cos. 

| 61560,78801 
78783 
78705 
78747 
78729 
78711 
78694 
78670 
78658 
78640 
78622 


61589 

! 61612 
I 61635 
J  61658 

!  61081 
!  61704 
I  61726 

II 61749 
61772 
61795 
61818  78604 


61841 
61864 
61887 
61909 
61932 


78586 
78568 
78550 
78532 
78514 


61355I78495 
61978  78478 


78460 
78442 
78424 
78405 
78387 
78369 
78351 
78333 
78315 
78297 
78279 
78261 
8243 
8225 
8200 
8188 
78170 
78152 
78134 
78116 
78098 
78079 
78001 
78043 
78025 
8007 
77988 
77970 
77952 
77934 
77916 
77897 
77879 
77801 
77843 
77824 
77808 
77788 
77769 
77751 
77733 
77715 
Tang.   |!  N.  coy.  N.sine 


62001 
1 1  62024 

!  1 62046 

J ;  02069 

62092 

62115 

62138 

62160 

62183 

62208 

62229 

62251 

62274 

62297 

62320 

62342 

62365 

62388 

62411 

62433 

62456 

62479 

625U2 

62524 

62547 

62570 

62592 

ij  62615 

I  i  62638 

j  1 62660 

;  J62683 

1 1 62706 

!  1 62728 

C2751 

j  1 62774 

j!  62796 

'J62819 

[62842 

|  62864 

| 62887 

!  1 62909 

62935 


51  Degrees. 


60 


Log.  Sines  and  Tangents.     (39°)    Natural  Sines. 


TABLE  II. 


10 

1  i 

12 

13 

14 

15 

16 

1? 

18 

19 

20 

21 

•2-2 

23 

■21 

26 

26 

•27 

28 

29 

30 

31 

32 

33 

34 

35 

36 

3? 

38 

39 

40 

41 

42 

43 

44 

15 

46 

47 

48 

49 

50 

51 

52 

53 

54 

65 

56 

57 

58 

59 

60 


9.798772 
799028 
799184 
799339 
799495 
799651 
799803 
799962 
800117 
800272 
800427 
9.800582 
800737 
800892 
801047 
801201 
801356 
801511 
801665 
801819 
801973 
3.802128 
802282 
802436 
802589 
802743 
802897 
803050 
803204 
803357 
803511 
>.  803664 
803817 
803970 
804123 
804276 
804428 
804581 
804734 
804886 
805039 
'.805191 
805343 
805495 
805547 
805799 
805951 
805103 
808254 
805405 
805557 
9.808709 
805860 
807011 
807163 
807314 
807465 
807615 
807766 
807917' 
803067 
Cosine. 


D.  10  I  Cosine. 


26.0 
26.0 
26.0 
25.9 
25.9 
25.9 
25.9 
25.9 
25.9 
25.8 
25.8 
25.8 
25.8 
25.8 
25.8 
25.8 
25.7 
25.7 
25.7 
25.7 
25.7 
25.7 
25.6 
25.6 
25.6 
25.6 
25.6 
25.6 
25.6 
25.5 
55.5 
25.5 
25.5 
25.5 
25-5 
25.4 
25.4 
25.4 
25.4 
25.4 
25.4 
25.4 
25.3 
25.3 
25.3 
25.3 
25.3 
25.3 
25.3 
25.2 
25.2 
25.2 
25.2 
25.2 
25.2 
25.2 
25.1 
25.1 
25.1 
25.1 


9.890503 
890400 
890298 
890195 
890093 
889990 
889888 
889783 
889682 
889579 
889477 

9.889374 
889271 
889168 
889054 
888961 
888858 
888755 
888651 
888548 
888444 

9.888341 
888237 
888134 
888030 
887926 
887822 
887718 
887614 
887510 
887406 

9.887302 
887198 
887093 
888989 
83.885 
886780 
886676 
836571 
886466 
886362 

9.888257 
888152 
886047 
885942 
885837 
885732 
885627 
885522 
885416 
885311 

9.885205 
885100 
884994 
884839 
884783 
884677 
884572 
881466 
884360 
884254 

Bine. 


D.  10 

17.0 
17.1 
17.1 
17.1 


17 

17 

!7 

17 

IV 

17 

17 

17.2 

17.2 

17.2 

17.2 

17.2 

17.2 

17.2 

17.2 

17.2 

17.3 

17.3 

17.3 

17.3 

17.3 

17.3 

17.3 

17.3 

17.3 

17.3 

17.4 

17.4 

17.4 

17.4 

17.4 

17.4 

17.4 

17.4 

17.4 

17.4 

17.5 

17.5 

17.5 

17.5 

17.5 

17.5 

17.5 

17.5 

17.5 

17.5 

17.6 

17.6 

17.6 

17.6 

17.6 

17.6 

17.6 

17.6 

17.6 

17.6 


Tans. 


9.903369 
903828 
903888 
909144 
909402 
909880 
909918 
910177 
910435 
910593 
910951 
9.911209 
911487 
911724 
911982 
912240 
912498 
912756 
913014 
913271 
913529 
9.913787 
914044 
914302 
914560 
914817 
915075 
915332 
915590 
915847 
916104 
916362 
916619 
916877 
917134 
917391 
917648 
917905 
918163 
918420 
918677 
9.918934 
919191 
919448 
919705 
919962 
920219 
920476 
920733 
920990 
921247 
9.921503 
921760 
922017 
922274 
922530 
922787 
923044 
923300 
923557 
923813 


Co  tang. 


D.  W 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

43.0 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.9 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.8 

42.7 


Co  cam 


10.091631 

091372 

091114 

090856 

090598 

090340 

090082 

089823 

089565 

089307 

089049 

10.088791 

088533 

088276 

088018 

'  087760 

087502 

087244 

086986 

086729 

086471 

10-086213 

085956 

085698 

085440 

085183 

034925 

084668 

084410 

084153 

083896 

10- 083638 

083381 

083123 

082866 

082609 

082352 

082095 

081837 

081580 

081323 

10-081066 

080809 

080552 

030295 

080038 

079781 

079524 

079267 

079010!! 

0787531; 

10.078497|| 

07824011 

077983! 

077726  j! 

077470J' 

077213 

076956 

076700 

076443 

076187 


62932 

(.2955 

6297? 

63000J 

63022 

63045 

63058 

63090 

63113 

63135 

63158 

93180 

63203 

63225 

63248 

63271 

63293 

63316 

63338 

63361 

63383 

63405 

63428 

63451 

63473 

63496 

63518 

63540 

63563 

63535 

63608 

63630 

63653 

63675 

63698 

63720 

63742 

63765 

63787 

63810 

63832 

63854 

6387  i 

63899 

63922 

63944 

63966 


77715 
77696 
77678 
77660 
77641 
77623 
77605 
77586 
77568 
77550 
77531 
77513 
77494 
77476 
77458 
77439 
77421 
77402 
77384 
77366 
77347 
77329 
77310 
77292 
77273 
255 
77236 
77218 
77199 
77181 
77162 
77144 
77125 
77107 
77088 
77070 
77051 
77033 
77014 
76996 
76977 
76959 
76940 
76921 
76903 
76884 
7o868 


63985176847 
64011  76828 


Tan!?. 


64033 
64056 
64078 
64100 
64123 
64145 
64167 
64190 
64212 
64234 
64256 
64279 
N.  coft 


76810 
76791 
76772 
76754 
76735 
76717 
76698 
76679 
76661 
76642 
76623 
76604 


.N.Hinc. 


50  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (40°)    Natural  Sines. 


Gl 


0 

1 

2 
3 
4 
5 
6 
7 
8 
9 
10 

11 
is 

13 
14 
15 
16 
17 
18 
19 
£20 
•21 
22 
■23 
•24 
•25 
86 
2? 
28 
29 
30 
31 
32 
33 
34 
86 
30 
37 
38 
39 
40 
41 
42 
43 
44 
45 

46 

47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
00 


Sine. 

9.803087 
808218 
808308 
808519 
808669 
808819 
808969 
809119 
809269 
809419 
809569 

9.809718 
809868 
810017 
810167 
810316 
810465 
810614 
810763 
810912 
811081 

9.811210 
811358 
811507 
811655 
811804 
811952 
812100 
812248 
812396 
812544 
.812692 
812840 
812988 
813135 
813283 
813430 
813578 
813725 
813872 
814019 
.814166 
814313 
814460 
814607 
814753 
814900 
815046 
815193 
815339 
815485 
.815631 
815778 
815924 
816069 
816215 
816361 
816507 
816652 
816798 
816943 
Cosine.  • 


D.  10" 


25.1 

25.1 

25.1 

25.0 

25.0 

25.0 

25.0 

25.0 

25.0 

24.9 

24.9 

24.9 

24.9 

24.9 

24.9 

24.8 

24.8 

24.8 

24.8 

24.8 

24.8 

24.8 

24.7 

24.7 

24.7 

24.7 

24.7 

24.7 

24.7 

24.6 

24.6 

24.6 

24.6 

24.6 

24.6 

24.6 

24.5 

24.5 

24 

24 

24 

24 

24 

24 

24 

24.4 

24.4 

24 

24 

24 

24 

24 

24 

24.3 

24.3 

24.3 

24.3 

24.2 

24.2 

24.2 


Cosine.  |D.  10" 


.884254 
884148 
884042 
883936 
883829 
883723 
883617 
883510 
883404 
883297 
883191 
,883084 
882977 
882871 
882764 
882657 
882550 
882443 
882336 
882229 
882121 
.882014 
881907 
881799 
881692 
881584 
881477 
881369 
881261 
881153 
881046 
.880938 
880830 
880722 
880613 
880505 
860397 
880289 
880180 
880072 
879963 
.879855 
879746 
879637 
879529 
879420 
879311 
879202 
879093 
878984 
878875 
.878766 
878656 
878547 
878438 
878328 
878219 
878109 
877999 
877890 
877780 
"Stile. 


7.7 
7.7 
7.7 
7.7 
7.7 
7.7 
7.7 


7.9 

7.9 

7.9 

7.9 

7.9 

7.9 

8.0 

8.0 

8.0 

8.0 

8.0 

8.0 

8.0 

8.0 

8.0 

8. 

8. 

8. 

8. 

8. 

8. 

8. 

8. 

8. 

8. 

8.2 

9. 


8 
8 

a 

8.2 
8.3 
8.3 
8.3 
8.3 


Tarn 


.923813 
924070 
924327 
924583 
924840 
925096 
925352 
925609 
925865 
926122 
926378 

.926634 
926890 
927147 
927403 
927659 
927915 
928171 
928427 
928683 
928940 

.929196 
929452 
929708 
929964 
930220 
930475 
930731 
930987 
931243 
931499 

'•931755 
932010 
932266 
932522 
932778 
933033 
933289 
933545 
933800 
934056 

.934311 
934567 
934823 
935078 
935333 
935589 
935844 
936100 
936355 
936610 

1.936866 
937121 
937376 
937632 
937887 
938142 
938398 
938653 
938908 
939163 

Cotang. 


D.  10' 


42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 
42.7 


42.7 
42.7 
42.7 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.6 
42.5 
42.5 
42.5 
42.5 
42.5 
42.5 
42.5 
42.5 
42.5 


Cotang.  hN.sine 


10.076187 
075930 
075673 
075417 
076160 
074904 
074648 
074391 
074135 
073878 
073622 

10.073366 
073110 
072853 
072597 
072341 
072085 
071829 
071573 
071317 
071060 

10.070804 
070548 
070292 
070036 
069780 
069525 
069269 
069013 
068767 
068501 

10.068245 
067990 
067734 
067478 
067222 
066967 
066711 
066455 
066200 
065944 

10.065689 
065433 
065177 
064922 
064667 
064411 
064156 
063900 
063645 
063390 

10.063134 
062879 
062624 
062368 
062113 
061858 
061602 
061347 
061092 
060837 


64279 
64301 
64323 
64346 
64368 
64390 
64412. 
64435 
64457 
64479 
64501 
64524 
64546 
64568 
64590 
64612 
64635 
64657 
64679 
64701 
64723 
64746 
64768 
64790 
64812 
64834 
64856 
64878 
64901 
64923 
64945 
64967 
.  64989 
1165011 
65033 
65055 
65077 
65100 
65122 
65144 
65166 
65188 
65210 
65232 
65254 
65276 
65298 
65320 
65342 
65364 
65386 
65408 
65430 
i 1 65452 
'65474 
v 65496 
i!  65518 
65540 
!!  65562 
|65§84 
I  656 06 


N,  cos. 

76604 
76586 

6567 
76548 
76530 
76511 
76492 
76473 
76455 
76436 
76417 
76398 
76380 
76361 
76342 
76323 

6304 
76286 

6267 
76248 
76229 
76210 
76192 
76173 
76154 
76135 
76116 
76097 
76078 
76059 
76041 
76022 
76003 
75984 
75965 
75946 
75927 
75908 
75889 
75870 
75851 
75832 
75813 
75794 
75775 
75/56 
75738 
76719 
75700 
75680 
75661 
75642 
75623 
75604 
75585 
75566 
75547 
75528 
75509 
75490 
75471 


Tang. 


N.  eos.  N.sine, 


49  Degrees. 


6:2 


Log.  Sines  and  Tangents.  (41°)  Natural  Sines. 


TABLE  IT. 


0 
1 
2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

•2-2 

23 

24 

25 

2o 

27 

28 

29 

30 

31 

32 

33 

31- 

35 

36 

3? 

38 

39 

40 

41 

42 

43 

44 

45 

46 

47 

48 

4!) 

50 

51 

52 

53 

54 

55 

50 

57 

58 

59 

60 


9.816943 
817088 
817233 
817379 
817524 
817668 
817813 
817958 
818103 
818247 
818392 

9.818536 
818681 
818825 
818969 
819113 
819257 
819401 
819545 
819689 
819832 

9.819976 
820120 
820263 
820403 
820550 
820893 
820836 
820979 
821122 
821265 

9.821407 
821550 
821693 
821835 
821977 
822120 
822262 
822404 
822546 
822688 

9.822830 
822972 
823114 
823255 
823397 
823539 
823680 
823821 
823963 
824104 

9.824245 
824386 
824527 
824668 
824808 
824949 
825090 
825230 
825371 
825511 
Cosine. 


U.  10 


24.2 

24.2 

24.2 

24.2 

24.1 

24.1 

24.1 

24.1 

24.1 

24.1 

24.1 

24.0 

24.0 

24.0 

24.0 

24.0 

24.0 

24.0 

23.9 

23.9 

23.9 

23.9 

23.9 

23.9 

23.9 

23.8 

23.8 

23.8 

23.8 

23.8 

23.8 

23.8 

23.8 

23.7 

23.7 

23.7 

23.7 

23.7 

23.7 

23.7 

23.6 

23.6 

23.6 

23.6 

23.6 

23.6 

23.6 

23.5 

23.5 

23.5 

23.5 

23.5 

23.5 

23.5 

23.4 

23.4 

23.4 

23.4 

23.4 

23.4 


Cosine. 

9.877780 
877670 
877560 
877450 
877340 
877230 
877120 
877010 
876899 
876789 
876678 

9.876568 
876457 
876347 
876236 
876125 
876014 
875904 
875793 
875682 
875571 

9.875459 
875348 
875237 
875126 
875014 
874903 
874791 
874680 
874568 
874456 

9.874344 
874232 
874121 
874009 
873896 
873784 
873672 
873560 
873448 
873335 

9.873223 
873110 
872998 
872885 
872772 
872659 
872547 
872434 
872321 
872208 

9.872095 
871981 
871868 
871755 
871641 
871528 
871414 
871301 
871187 
871073 


D.  10" 


Sine. 


18.3 

18.3 

18.3 

18.3 

18 

18 

18 

18 

18 

18 

18 

18.4 

18.4 

18.4 

18.5 

18.5 

18.5 

18.5 

18.5 

18.5 

18.5 

18.5 

18.5 

18.5 

18.6 

18.6 

18.6 

18.6 

18-.  6 

18.6 

18.6 

18.6 

18.7 

18.7 

18.7 

18.7 

18.7 

18.7 

18.7 

18.7 

18.7 

18.7 

18.8 

18.8 

18.8 

18.8 

18.8 

18.8 

18.8 

18.8 

18.8 

18.9 

18.9 

18.9 

18.9 

18.9 

18.9 

18.9 

18.9 

18.9 


Tan?:. 


939163 
939418 
939673 
939928 
940183 
940438 
940394 
940949 
941204 
941458 
941714 

9.941968 
942223 
942478 
942733 
942988 
943243 
943498 
943752 
944007 
944252 
944517 
944771 
945026 
945281 
945535 
945790 
948045 
946299 
948554 
946803 

9.947053 
947318 
947572 
947826 
948081 
948336 
948590 
948844 
949099 
949353 

9.949607 
949862 
950116 
950370 
950625 
950879 
951133 
951388 
951642 
951898 

1.952150 
952405 
952659 
952913 
953167 
953421 
953675 
953929 
954183 
951437 
Cotang. 


D.  10" 


42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.5 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42 

42 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 

42.4 


Cotang.  '  N.  sine. 


10.0308371!  65606 
060582  1 165628 
080327  165650 
060072  i 65672 
059817  1 165694 
0595621,65716 
059306  |!  65738 


059051  !  65759  75337 


X.  cos. 

754 

75452 

75433 

75414 

75395 

75375 

75356 


058796'!  65781^ 
058542!  65803 
058286  l|  65825 
10.0580321  65847 
057777:  65869 
057522  i  65891 
057267  ,65913 
057012  j!  65935 
056757  !'  65956 


42 

42 

42 

42 

42 

42 

42 

42 

42.3 

42.3 

42.3 

42.3 


056502  165978 
056248  66000 
055993  66022 
055738  ','66044 

10.055483  66066 
055229J  66088 
054974  66109 
054719  63131 
054465  '66153 
054210  66175 
053955  6819 
053701  |  66218 
053446  '66240 
053192  : 66262 

10.052937!  66284 
052682 !  68308 
052428!  1 6632 
052174  166349 
051919i  166371 
051664  |!  66393 
051410!  66414 


051156 
050901 
050647 

10.050393 
050138 
049884 
049630 
049375 
049121 
048867 
048612 
048358 
048104 ! 

10.047850' 
047595  I 
047341 i 
047087 : 
046833 : 
046579 
046325 
046071 
045817 
045563  ! 
Tana. 


! 66436 

66458 
! 66480 
66501 
; 66523 
66545 
66566 
! 66588 
66610 
66832 
6665 
66675 
66697 
66718 
66740 
66762 
66783 
66805 
66827 
66848 
66870 
66891 
66913 

N.  cos 


75318 
75299 
75280 
75261 
75241 
75222 
75203 
75184 
75165 
75146 
75126 
75107 
75083 
75069 
75050 
75030 
75011 
74992 
74973 
74353 
74934 
74915 
74896 
74876 
74857 
74838 
74818 
74799 
74780 
74760 
74741 
74722 
74703 
74683 
74663 
4644 
4625 
74605 
74586 
74567 
74548 
74522 
74509 
74489 
74470 
74451 
74431 
74412 
74392 
74373 
74353 
74334 
74314 


X.sine 


60 

59 
58 
57 
56 
55 
54 
53 
52 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
36 
84 
38 
82 
31 
30 
29 
28 
27 
26 
25 
24 
23 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0 


48  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (42°)    Natural  Sines. 


63 


Sine.       D.  10'      Cosine.      D.  10"|      Tang. 


23.4 
23.3 


23.3 


825511 
825651 
825791 
835931  \f%% 

826971  -°-J 
826211 
826351 
826491 
826631 
826/70 
826910 
.827049 
827189 
827328 
827467 
827606 
827745 
827834 
828023 
828162 
828301 
.828439 
828578 
828716 
828855 
828993 
829131 
829269 
829407 
829545 
829683 
.829821 
829959 
830097 
830234 
830372 
830509 
830846  I 
830784  i 
830921  | 
831058  ! 
.831195, 
831332 
8314691 
831606  J 
831742 
831879 
832015 
832152 
832288 
832425 
832561  ; 
832697  ; 
832833 
832969 
833105 
833241  li-l 
833377  f-b 
833512  H -° 
833648 
833783 


23.3 

2;. 3 

23.3 
.3.3 
23.2 
23.2 
23.2 
23.2 
23.2 
23.2 
23.2 
23.2 
23.1 
23.1 
23.1 
23.1 
23.1 
23.1 
23.1 
23.0 
23.0 
23.0 
23.0 
23.0 
23.0 
23.0 
22.9 
22.9 
22.9 
J22.9 
|22.9 
'22.9 
22.9 
22.9 
22.8 
22.8 
22.8 
22.8 
22.8 
22.8 
22.8 
22.8 
22.7 
22.7 
22.7 
22.7 
22.7 
22.7 
22.7 
22.6 


22.6 
22.6 


Cosine 


.871073 
870960 
870846 
870732 
870318 
870504 

'870390 
870276 
870161 
870047 
839933 

.869818 
859704 
869589 
869474 
889360 
869245 
869130 
869015 
8J8900 
868785 

.868670 
868555 
868440 
868324 
868209 
868093 
857978 
867862 
867747 
867631 

.867515 
867399 
867283 
867167 
867051 
866935 
866819 
866703 
866586 
866470 

.866353 
866237 
866120 
866004 
865887 
865770 
865653 
865536 
865419 
865302 

1.865185 
865068 
864950 
864833 
864716 
864598 
864481 
864363 
864245 
864127 


Sine. 


19.0 
19.0 
19.0 
19.0 
19.0 
19.0 
19.0 
19.0 
19.0 
19.1 
19.1 
19.1 
19.1 
19.1 
19.1 
19.1 
19.1 
19.1 
19.2 
19.2 
19.2 
19.2 
19.2 
19.2 
19.2 
19.2 
19.2 
19.3 
19.3 
19.3 
19.3 
19.3 
19.3 
19.3 
19.3 
19.3 
19.4 
19.4 
19.4 
19.4 
19.4 
19.4 
19.4 
19.4 
19.5 
19.5 
19.5 
19.5 
19.5 
19.5 
19.5 
19.5 
19.5 
19.5 
19.6 
19.6 
19.6 
19.6 
19.6 
19.6 


.954437 
954691 
954945 
955200 
955454 
955707 
955961 
956215 
956469 
956723 
956977 

.957231 
957485 
957739 
957993 
958246 
-958500 
958754 


959262 
959516 

.959769 
960023 
960277 
960531 
960784 
961038 
961291 
961545 
961799 
962052 

.962306 
962560 
962813 
963067 
963320 
963574 
963827 
964081 
964335 
964588 

.964842 
965095 
965349 
965602 
965855 
966109 
966362 
966616 
966869 
967123 

.967376 
967629 
967883 
968136 
968389 
968643 
968896 
969149 
969403 
969656 


Cotan" 


D.  10" 


42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42 

42 

42 

4-2 

42 

42 

42 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.3 

42.2 

42.2 

42.2 

42 

42 

42 

42 

42 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 

42.2 


Cotang.  j  N.  sine. IN.  cos. 


10.045563 
045309 
045055 
044800 
044546 
044293 
044039 
043785 
043531 
043277 
043023 

10.042769 
042515 
042261 
042007 
041754 
041500 
041246 
040992 
040738 
040484 

10.040231 
039977 
039723 
039469 
039216 
038962 
038709 
038455 
038201 
037948 

10.037694 
037440 
037187 
036933 
036680 
036426 
036173 
035919 
035665 
035412 

10.035158 
034905 
034651 
034398 
034145 
033891 
033638 
033384 
033131 
032877 

10.032624 
032371 
032117 
031864 
031611 
031357 
031104 
030851 
030597 
030344 


j]  66913174314 
i!  66935174295 
j!  66956 174276 
166978174256 
i!  66999  74237 


67021 


! 

! 67043 
! 67064 
1 67086 
67107 
67129 
67151 
67172 
67194 


74217 
74198 
74178 
74159 
74139 
74120 
74100 
74080 
.74061 
67215  !74041 
67237 174022 
67258  j  74002 
67280173983 
67301  73963 


67323 
! 67344 
I 67366 

! 67387 

! 67409 

I ! 67430 

j ! 67452 

| '67473 

I 67495 

167516 

! 67538 

j  67559 

,167580 

!  167602 

;!  67623 

167645 

1 1  67666 

1 1 67688 

I S67709 

|  i 67730 

| | 67752 

67773 

67795 

67816 

67837 

67859 

67880 

67901 

67923 

67944 

67965 

67987 

68008 

68029 

68051 

68072 

68093 

68115 

68136 

68157 

68179 

68200 


Tang.   ||  N.  cos.  N.sine 


73944 
73924 
73904 
73885 
73865 
73846 
73826 
73806 
73787 
73767 
73747 
73728 
73708 
73688 
73669 
73649 
73629 
73610 
73590 
73570 
73551 
73531 
73511 
73491 
73472 
73452 
73432 
73413 
73393 
73373 
73E53 
73333 
73314 
73294 
73274 
73254 
73234 
73215 
73195 
73175 
73155 
73135 


47  Degrees. 


64 


Log.  Sines  and  Tangents.  (43°)  Natural  Sines. 


TABLE  II. 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

30 

•21 

22 

23 

24 

25 

36 

27 

28 

29 

30 

31 

32 

33 

34 

35 

36 

37 

38 

39 

40 

41 

42 

43 

44 

45 

40 

47 

48 

49 

50 

51 

52 

53 

54 

55 

5G 

57 

58 

59 

00 


Sine. 

9.833783 
833919 
834054 
834189 
834325 
834460 
834595 
834730 
834865 
834999 
835134 
835269 
835403 
835538 
835672 
835807 
835941 
836075 
836209 
836343 
836477 
836611 
836745 
836878 
837012 
837146 
837279 
837412 
837546 
837679 
837812 
9.837945 
838078 
838211 
838344 
838477 
838610 
838742 
838875 
839007 
839140 

.839272 
839404 
839536 
839668 
839800 
839932 
840084 
840190 
840328 
840459 

.840591 
840722 
840854 
840985 
841116 
841247 
841378 
841509 
841640 
841771 

O  sine. 


D.  10" 


22.6 
22.5 
22.5 
22.5 
22.5 
22.5 
22.5 
22.5 
22.5 
22.4 
22.4 
22.4 
22.4 
22.4 
22.4 
22.4 
22.4 


Cosine. 


22 

22 

22 

22 

22 

22 

22.3 

22.2 

22.2 

22.2 

22.2 

22.2 

22.2 

22.2 

22.2 

22.1 

22.1 

22.1 

22.1 

22.1 

22.1 

22.1 

22.1 

22.0 

22.0 

22.0 

22.0 

22.0 

22.0 

22.0 

21.9 

21.9 

21.9 

21.9 

21.9 

21.9 

21.9 

21.9 

21.8 

21.8 

21.8 

21.8 

21.8 


>.  864127 
864010 
863892 
863774 
863656 
863538 
863419 
863301 
863183 
863064 
862946 
.882827 
862709 
862590 
862471 
862353 
862234 
•862115 
861998 
861877 
861758 
.861638 
851519 
861400 
861280 
861161 
861041 
860922 
860802 
860682 
860562 
.860442 
860322 
860202 
860082 
859962 
859842 
859721 
859601 
859480 
859360 
859239 
859119 
858998 
858877 
858756 
858635 
858514 
858393 
858272 
858151 
,858029 
857903 
857786 
857665 
857543 
857422 
857300 
857178 
357056 
855934 
"line. 


D.  10" 


19 
19 
19 
19 
19 
19 
19 
19 
19 
19 
19 
19 
19 
19 
19. 
19, 
19, 
19, 
19, 
19, 
19, 
19. 
19, 
19. 
19. 
19, 
19. 
19. 
19. 
20. 
20. 
20. 
20. 
20. 
■JO. 
20 
20. 
20. 
20. 
20. 
20. 
20. 

20. 
20. 
20. 
20. 
20. 
20. 
2D. 
20. 
20. 
20. 
20. 
20. 
20. 
20. 
20. 
20. 
20. 
20. 


Tang.  |D.  10" 


9.969656 
969909 
970162 
970416 
970669 
970922 
971175 
971429 
971682 
971935 
972188 

5.972441 
972694 
972948 
973201 
973454 
973707 
973960 
974213 
974466 
974719 

). 974973 
975226 
975479 
975732 
975985 
976238 
976491 
976744 
976997 i 
977250 

).  977503 
977756 
978009 
978262 
978515 
978768 
979021 
979274 
979527 
979780 

1.980033 
980286 
980538 
980791 
981044 
981297 
981550 
981803 
982056 
982309 

'.982562 
982814 
983067 
933320 
983573 
983826 
934079 
984331 
984584 
984837 


42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42 

42, 

42, 

42, 

42, 

42. 

12. 

42. 

42. 

42. 

42. 

42 

42 

42 

42 

42 

42 

42 

42 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 

42. 


Cotang.  |  |N  .sine.  N 

68200 
68221 
68242 
68264 
68285 
68306 
68327 
68349 
68370 
68391 
68412 
68434 
68455 
68476 
68497 
68518 


0  10.030344 
030091 
029838 
029584 
029331 
029078 
028825 
028571 
028318 
028065 
027812 

10.027559 
027306 
027052 
026799 
026546 
026293 
026040 
025787 
025534 
025281 

10.025027 
024774 
024521 
02426a 
024015 
023762 
023509 
023256 
023003 
022750 | 

10.022497 
022244 
021991 
021738 
021485 
021232 
020979 
020726 


73135 
73116 
73096 
73076 
73056 
73036 
73016 
72996 
72976 
72957 
72937 
2917 
72897 
72877 
72857 
72837 
68539172817 
68561 J72797 


68582 
68603 
68624 
68645 
68666 
68688 


72777 
72757 
72737 
72717 
72697 
7 


68709J72657 
6873072637 


Cotany 


68772  72597 
68793172577 
6881472557 
68835  72537 
68857  72517 
6887872497 
6889972477 
68920  72457 
6894172437 
68962  72417 
68983,72397 
6900472377 
020473  69025  72357 
020220!  6904672337 
10.019967'  69087  72317 
019714  69088  72297 
019462  J  69109  72277 
019209  !  1 69130  72257 
018956^6915172236 
018703  I  6917272216 
018450  |69193  72196 
018197:  69214  72176 
017944!  69235  72156 
017691  69256  72136 
10.017438  i 69277  72116 
017186  16929872095 
016933  :  69319  72075 
016680,1  69340  72055 
016427  6936172035 
016174  69382  72015 
015921  169403  71995 
015669  69424  71974 
015416  69445  71954 
015163  6946671934 


Tan":. 


N.  cos. |N. sine. 


60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 
49 
48 
47 
46 
45 
44 
43 
42 
41 
40 
39 
38 
37 
36 
35 
34 
33 
32 
31 
30 
39 
28 
27 
26 
26 
24 
23 
33 
21 
20 
19 
IS 
17 
16 
15 
14 
13 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0 


46  Degrees. 


TABLE  II. 


Log.  Sines  and  Tangents.    (44°)    Natural  Sines. 


05 


Sim;. 

). 841771 
841902 
842033 
842163 
842294 
842424 
842555 
842685 
842815 
842946 
843076 

•  .843206 
843336 
843466 
843595 
843725 
843855 
843984 
844114 
844243 
844372 

1.844502 
844631 
844760 
844889 
845018 
845147 
845276 
845405 
845533 
845662 

.845790 
845919 
846047 
846175 
846304 
846432 
846560 
846688 
846816 
846944 

.847071 
847199 
847327 
847454 
847582 
847709 
847836 
847964 
848091 
848218 

.848345 
848472 
848599 
848726 
848852 
848979 
849106 
849232 
849359 
849485 

Cosine. 


D.  10"  Cosine. 


,856934 
856812 
856690 
856568 
856446 
856323 
856201 
856078 
855956 
855833 
855711 
,855588 
855465 
855342, 
855219 
855096 
854973 
854850 
854727 
854603 
854480 
854356 
854233 
854109 
853986 
853862 
853738 
853614 
853490 
853366 
853242 
853118 
852994 
852869 
852745 
852620 
852496 
852371 
852247 
852122 
851997 
851872 
851747 
851622 
851497 
851372 
851246 
851121 
850996 
850870 
850745 
9.850619 
850493 
850368 
850242 
850116 
849990 
849864 
849738 
849611 
849485 
Sine.  • 


D.  10" 


Till!'. 


984837 
985090 
985343 
985596 
985848 
986101 
986354 
986607 
986860 
987112 
987365 

9.987618 
987871 
988123 
988376 
988629 
988882 
989134 
989387 
989640 
989893 
990145 
990398 
990651 
990903 
991156 
991409 
991662 
991914 
992167 
992420 
992672 
992925 
993178 
993430 
993683 
993936 
994189 
994441 
994694 
994947 
995199 
995452 
995705 
995957 
996210 
996463 
996715 
996968 
997221 
997473 

9.997726 
997979 
998231 
998484 
998737 


999242 

999495 

999748 

10.000000 


Cotan» 


D>10" 

42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 
42.1 


42 

42 

42 

42 

42 

42.1 

42.1 


42.1 


Cotang. 


N.  sine.  N. 


10.015163 

014910 

014657 

014404 

014152 

013899 

013646 

013393 

013140 

012888 

012635  I 
10.012382 

012129 

011877 

011624 

011371 

011118 

010866 

010613 

010360 

010107 
10.009855 

009602 

009349 

009097 

008844 

008591 

008338 

008086 

007833 

007580 
10-007328 

007075 

006822 

006570 

006317 

006064 

005811 

005559 

005306 

005053  1 70298 
10-004801  |  70319 

004548  [  I  70339 

004295  I  70360 

004043  I  70381 

003790!  70401 

003537  I  70422 


69466 

6948 

69503 

69529 

69549 

69570 

69591 

69612 

69633 

69654 

69675 

69696 

697 

69737 

69758 

69779 

69800 

69821 

69842 

69862 

69883 

69904 

69925 

69946 

69966 

69987 

70008 

70029 

70049 

70070 

70091 

70112 

70132 

70153 

70174 

70195 

70215 

70236 

70257 

70277 


003285 
003032 
002779 
002527 
10-002274 
002021 
001769 
001516 
001263 
001011 
000758 
000505 
000253 
000000 


| 70443 
70463 
70484 
70505 
70525 
70546 
70567 
70587 
70608 
70628 
70649 
70670 
70690 
70711 


Tarn 


71934 
71914 
71894 
71873 
71853 
71833 
71813 
71792 
71772 
71752 
71732 
71711 
71691 
71671 
71650 
71630 
71610 
71590 
71569 
71549 
71529 
71508 
71488 
71468 
71447 
71427 
71407 
71386 
71366 
71345 
71325 
71305 
71284 
71264 
71243 
71223 
71203 
71182 
71162 
71141 
71121 
71100 
71080 
7.1059 
71039 
71019 
70998 
0978 
70957 
0937 
70916 
70896 
70875 
70855 
0834 
70813 
0793 
0772 
0752 
0731 
70711 
N.  cos.  N.sine. 


45  Decrees. 


66 

LOGARITHMS 

TABLE  III. 

LOGARITHMS   OF    NUMBERS. 

From  1  to  200, 

INCLUDING 

TWELVE  DECIMAL 

PLACES. 

N. 

Log. 

|K. 

Log. 

1  N. 

Log. 

1 

oooooo  oooooo 

41 

612783  856720  . 
623249  290398 
633468  455580 

\     81 

908485  018879 

2 

301029  995864 

42 

82 

913813  852384 

3 

477121  254720 

43 

83 

919078  092376 

4 

602059  991328 

44 

643452  676486 

84 

924279  286062 

5 

698970  004336 

46 

653212  513775 

85 

929418  925714 

6 

778151  250384 

46 

662757  831682 

86 

934498  451244 

7 

845098  010014 

47 

672097  857926 

87 

939519  252619 

8 

*903089  986992 

48 

681241  237376 

88 

944482  672150 

9 

954242  509439 

49 

690193  080028 

89 

949390  006645 

10 

Same  as  to  1. 

60 

Same  as  to  5. 

90 

Same  as  to  9. 

11 

041392  685158 

51 

707570  176098 

91 

959041  392321 

12 

079181  246018 

52 

716003  343635 

92 

963787  827346 

13 

113943  352307 

63 

724275  869601 

93 

968482  948554 

14 

146128  035678 

64 

732393  759823 

94 

973127  853600 

15 

176091  259056 

55 

740382  689494 

95 

977723  605889 

16 

204119  982656 

56 

748188  02700S 

95 

982271  233040 

17 

230448  921378 

57 

755874  855672 

97 

986771  734266 

18 

255272  505103 

68 

763427  993563 

98 

991226  075692 

19 

278753  600953 

69 

770852  011642 

99 

995635  194598 

20 

Same  as  to  2. 

60 

Same  as  to  6 

100 

Same  as  to  10. 

21 

322219  2947 

61 

785329  835011 

101 

004321  373783 

22 

342422  680822 

62 

792391  699498 

102 

008600  171762 

23 

361727  836018 

63 

799340  549453 

103 

012837  224705 

24 

380211  241712 

64 

808179  973984 

104 

017033  339299 

25 

397940  008672 

65 

812913  356643 

105 

021189  299070 

26 

414973  347971 

66 

819543  935542 

103 

025305  865265 

27 

431363  764159 

67 

826074  802701 

107 

029383  777685 

28 

447158  031342 

68 

832508  912706 

108 

033423  755487 

29 

462397  997899 

69 

838849  090737 

109 

037426  497941 

30 

St- me  as  to  3. 

70 

Same  as  to  7. 

110 

Same  as  to  11. 

31 

491361  693834 

71 

851258  348719 

111 

045322  978787 

32 

505149  978320 

72 

857332  496431 

112 

049218  022670 

33 

518513  939878 

73 

863322  860120 

113 

053078  443483 

34 

531478  917042 

74 

869231  719731 

114 

056904  851336 

35 

544068  044350 

75 

875061  263392 

115 

060397  840354 

36 

556302  500767 

76 

880813  592281 

116 

054457  989227 

37 

568201  724067 

77 

886490  725172 

117 

058185  861746 

38 

579783  596617 

78 

892094  602690 

118 

071882  007306 

39 

591054  607026 

79 

897627  091290 

119 

075546  961393 

40 

Same  as  to  4. 

80 

Same  as  to  8. 

120 

Same  as  to  12. 

OF    NUMBERS. 


67 


Is. 

Log._ 

082785  370316 

N. 

Log. 

N.  , 

Log 

121 

148 

170261  715395 

175 

243038  048686 

122 

086359  830875 

149 

173186  268412 

176 

245512  667814 

123 

08990o  111439 

150 

176091  259056 

177 

247973  266362 

124 

093421  685162 

151 

178976  947293 

178 

250420  002309 

125 

096910  013008 

;  152 

181843  587945 

179 

25:2853  030980 

126 

100370  545118 

153 

184691  430818 

180 

255272  505103 

127 

103803  720956 

1  154 

187520  720836 

181 

257678  574869 

128 

107209  969648 

!  155 

190331  698170 

182 

260071  387985 

129 

110589  710-299 

156 

193124  588354 

183 

262451  089730 

130 

Same  as  to  13. 

157 

195899  652409 

184 

264817  823010 

131 

117271  295656 

158 

198857  086954  i 

185 

267171  728403 

132 

120573  931206 

159 

201397  124320 

186 

269512  944218 

133 

123851  640967 

160 

204119  982656 

187 

271841  606536 

134 

127104  798365 

161 

208825  876032 

188 

274157  849284 

135 

130333  768495 

162 

209515  014543 

139 

276461  804173 

136 

133538  908370 

163 

212187  604404 

190 

278753  600953 

137 

136720  567156 

164 

214843  848048 

191 

281033  367248 

138 

139879  086401 

165 

217483  944214 

192 

283301  228704 

139 

143014  800254 

166 

220108  088040 

193 

285557  309008 

140 

146128  035678 

167 

222716  471148 

194 

287801  729930 

141 

149219  112655 

163 

225309  281726 

195 

290034  611362 

142 

152288  3443S3 

169 

227886  704614 

196 

292256  071356 

143 

155336  037465 

170 

230448  921378 

197 

294466  226162 

144 

158362  492095 

171 

232996  110392 

198 

296665  190262 

145 

161368  002235 

172 

235528  446908 

199 

298853  076410 

146 

164352  855784 

173 

238046  103129 

147 

167317  334748 

174 

240549  248283 

LOGARITHMS  OF  THE  PRIME   NUMBERS 

From  200  to   1543, 

INCLUDING   TWELVE   DECIMAL  PLACES. 


N. 


201 
203 
207 
209 
211 

223 
227 
229 
233 
239 

241 
251 
257 
263 
269 

271 


Loi 


303196  057420 
307496  037913 
315970  345457 
320146  286111 
324282  455298 

348304  863048 
356025  857193 
359835  482340 
367355  921020 
378397  900948 

3820T7  042575 
399873  721481 
409933  123331 
419955  748490 
429752  280002 

432989  290874 


N, 


277 
281 
283 
293 
307 

311 
313 
317 
331 
337 

347 
349 
353 
359 
367 

373 


442479  769064 
448708  319905 
451786  435524 
466867  620D54 
487138  375477 

492760  389027 
495544  337546 
501059  262218 
519827  993776 
527629  900871 

540329  474791 
542825  426959 
r 47774  705388 
555094  448578 
664666  004252 

571703  831809 


379 

383 
389 
397 
401 

409 
419 
421 
431 
433 

439 

443 
449 
457 
461 

463 


Lo^. 


578839  209968 
583198  773968 
589949  601326 
598790  508763 
603144  372020 

611723  308007 
622214  U22966 
624282  095836 
634477  270161 
636487  896353 

642424  520242 
646403  726223 
652246  341003 
659916  200070 
663700  925390 

665580  991018 


-  - 
68 

LOGARITHMS 

N. 

Log. 

N. 

Log. 

1171 

Log. 
uo»566  fc950:2 

467 

6J9blo  briU5b6 

"821 

914343  157119 

479 

680335  513414 

823 

915399  835212 

1181 

0/2249  807613 

487 

68/528  961215 

827 

917505  509553 

1137 

074450  718955 

491 

691081  492123 

829 

918554  530550 

1193 

076640  443670 

499 

69810 0  545623 

8S9 

923761  960829 

1201 

0.9543  007385 

503 

701567  985056 

853 

930949  031168 

1213 

083830  800345 

509 

706717  782337 

857 

932980  821923 

1217 

085290  678210 

521 

716S37  723300 

859 

933993  163331 

1223 

087426  458017 

623 

718501  688867 

863 

936010  795715 

1229 

089551  882866 

641 

733197  255107 

877 

942999  593356 

1231 

090258  052912 

547 

737987  326333 

881 

944975  908412 

1237 

092369  699609 

657 

745855  195174 

883 

945960  703578 

1249 

096562  438356 

563 

750508  394851 

887 

947923  619832 

1259 

100025  729204 

669 

755112  26639 > 

907 

957607  287060 

1277 

103190  896808 

571 

756636  1Q8213 

911 

959518  376973 

1279 

106870  542480 

577 

761175  813156 

919 

963316  511386 

1283 

108226  656362 

687 

768638  101248 

929 

968015  713994 

1289 

110252  917337 

593 

773U54  693364 

937 

971739  590888 

1291 

110926  242517 

599 

777426  822389 

941 

973589  623427 

1297 

112939  986066 

601 

778874  472002 

947 

976349  979003 

1301 

114277  296540 

607 

783138  691075 

953 

979092  900838 

1303 

114944  415712 

613 

787460  474618 

967 

985426  474083 

1307 

116275  587564 

617 

790285  164033 

9/1 

987219  229908 

1319 

120244  795568 

619 

791690  649020 

977 

989894  563719 

1321 

120902  817604 

631 

800029  359244 

9£3 

992553  617832 

1327 

122870  922849 

641 

806858  029519 

991 

996073  654485 

1361 

133858  125188 

643 

808210  972924 

997 

998695  158312 

1367 

135768  514554 

647 

810904  280669 

1009 

003891  166237 

1373 

137670  537223 

653 

814913  181275 

1013 

005609  445360 

1381 

140193  678544 

659 

818885  414594P 

1019 

008174  1840J6 

1399 

145817  714122 

661 

810201  459486 

1021 

009025  742087 

1409 

148910  994096 

673 

828015  064224 

1031 

013258  665284 

1423 

153204  896557 

677 

830588  668685 

1033 

014100  321520 

1427 

154424  012366 

683 

834420  703682 

1039 

016615  547557 

1429 

155032  228774 

691 

839478  047374 

1049 

020775  488194 

1433 

156246  402184 

701 

845718  017967 

1051 

021602  716028 

1439 

158060  793919 

709 

850646  235183 

1061 

025715  383901 

1447 

160468  531109 

719 

856728  890383 

1063 

026533  264523 

1451 

161667  412427 

727 

861534  410859 

1069 

028977  705209 

1453 

162265  614286 

733 

865103  974742 

1087 

036229  544036 

1459 

164055  291883 

739 

868644  488395 

1091 

037824  750588 

1471 

167612  672629 

743 

870988  813761 

1093 

038620  161950 

1481 

170555  058512 

751 

855639  937004 

1097 

040206  627575 

1483 

171141  151014 

757 

879095  879500 

1103 

042595  512440 

1487 

172310  968489 

761 

881384  656771 

1109 

044931  546119 

1489 

172894  731332 

!   769 

885926  339801 

1117 

048053  173116 

1493 

174059  807708 

773 

888179  493918 

1123 

050379  756261 

1499 

175801  632866 

787 

895974  732359 

1129 

052693  941926 

1511 

179264  464329 

797 

901458  321396 

1151 

031075  323630 

1523 

182699  903324 

809 

907948  521612 

1153 

061829  307295 

1531 

184975  190807 

811 

909020  854211 

1163 

065579  714728 

1543 

188365  926053 

OF    NUMBERS 


AUXILIARY    LOGARITHMS, 


1ST. 


1.009 
1.008 
1.007 
1.006 
1 .  005 
1.004 
1.003 
1.002 
1.001 


Log. 

1    ff. 

003891166237  >, 

1.0009 

003460532110 

1.0008 

003029470554 

1.0007 

002598080685 

1.0003 

0021660ol756 

\a 

1.0005 

001733712775 

1.0004 

001300933020 

1.0003 

000867721529 

1.0002 

000434077479  J 

1.0001 

Lost. 


000390689248 
000347296C84 
000303899784 
000260-198547 
000217092970 
000173683057 
000130268804 
000086850211 
000043427277 


c 


N. 

1 

.00009 

1 

.00008 

1 

.00007 

1 

.00006 

l 

.00005 

1 

.00004 

1 

.00003 

1 

.00002 

1 

00001 

Log- 


000039083266 
000034740691 
000030398072 
000026055410 
000021712704 
000017371430 
000013028638 
000008685802 
000004342923 


L 

i 

1ST.   f 
.000009 

i 

. 000008 

i 

.000007 

i 

.000006 

i 

.000005 

i 

.000004 

i 

000003 

i 

000002 

i 

000001 

Log. 


000003908628 
000003474338 
000003040047 
000002605756 
000002171464 
000001737173 
000001302880 
000000868587 
000000434294 


1.00000001 

1.000000001 

1.0000000001 


Log. 


000000043429 
000000004343 
000000000434 
000000000043 


(n) 
(o) 
(P) 
(q) 


»i=0.4342944819        log.  —1.637784298. 

By  the  preceding  tables  —  and  the  auxiliaries  A,  B,  and 
C,  we  can  find  the  logarithm  of  any  number,  true  to  at  leap.t 
ten  decimal  places. 

But  some  may  prefer  to  use  the  following  direct  formula, 
which  may  be  found  in  any  of  the  standard  works  on  algebra: 

Log.  (2-j-l)=log.2+0.8685889638^_J-  \ 

The  result  will  be  true  to  twelve  decimal  places,  if  z  be 
over  2000. 

The  log.  of  composite  numbers  can  be  determined  by  the 
combination  of  logarithms,  already  in  the  table,  and  the  prime 
numbers  from  the  formula. 

Thus,  the  number  3083  is  a  prime  number,  find  its  loga- 
rithm. 

We  first  find  the  log.  of  the  number  3082.  By  factoring, 
we  discover  that  this  is  the  product  of  46  into  67. 


70 


NUMBERS 


Log.  46, 
Log.  67, 

Log.  3082 
Log.  3083=3.4888326343- 


1.6627578316 
1.8260748027 

3.4888326343 

0.8685889638 


6165 


NUMBERS  AND  THEIR  LOGARITHMS, 

OFTEN    USED    IN    COMPUTATIONS. 


Log. 
3.14159265    0.4971499 


Circumference  of  a  circle  to  dia.  1 ) 

Surface  of  a  sphere  to  diameter  1 

Area  of  a  circle  to  radius  1 

Area  of  a  circle  to  diameter  1  =;   .7853982  —1.8950899 

Capacity  of  a  sphere  to  diameter  1    =   .5235988—1.7189986 

Capacity  of  a  sphere  to  radius  1     =4.1887902       0.6220886 


I- 


Arc  of  any  circle  equal  to  the  radius  =57°29578  1.7581226 
Arc  equal  to  radius  expressed  in  sec.  =206264"8  5.3144251 
Length  of  a  degree,  (radius  unity)  =  .01745329  —2.2418773 

12  hours  expressed  in  seconds,      =    43200  4.6354837 

Complement  of  the  same,       =0.00002315  —5.3645163 

360  degrees  expressed  in  seconds,  =   1296000  6.1 126050 


•  A  gallon  of  distilled  water,  when  the  temperature  is  62 
Fahrenheit,  and  Barometer  30  inches,  is  277.r20-VV  cubi< 
inches. 


10 


7277.274=16.651542  nearly. 


4 
■I 


277.274 
.775398 


=  18.78925284 


7231=15.198684. 


7282=16.792855. 


=  18.948708. 


.785398 

The  French  Metre=3.2808992,  English  feet  linear  mea- 
sure, =39.3707904  inches,  the  length  of  a  pendulum  vi- 
brating seconds. 


^        OF  THE        ^j\ 

UNIVERSITY  ) 


/ 


J 


ryiM4tf^ 


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